Question about proof from a guy with a highschool education

[reenmachine]

Hi everyone , I was trying to understand what a math proof is or more specifically how it is presented.

Suppose I want to prove that if A , B and C are real numbers and that (A + B) = C , then (A - B) = (C - 2B).

How would I present a proof for this? Would this be good?:

If (A + B) = C , then A = (C - B)
If (A - B) = (C - 2B) , then (A - B) + B = (C - 2B) + B (which gives us A = (C - B))
As previously demonstrated , if A = (C - B) , then (A + B) = C , therefore proving that if A , B and C are real numbers and that (A + B) = C , then (A - B) = (C - 2B).
I am just a newbie trying to learn so go easy on me.I understand this is a very short example.Any random thoughts on mathematical proofs will be appreciated.

 

[micromass]

Since you are wanting to learn proofs, I think you might benefit from me being very criticizing.

Hi everyone , I was trying to understand what a math proof is or more specifically how it is presented.

Suppose I want to prove that if A , B and C are real numbers and that (A + B) = C , then (A - B) = (C - 2B).

A notation issue. You don't write brackets around two expressions which are related by an equality. So something like (A+B) = C is not acceptable. You should write A+B = C.
Brackets are only useful when you got something like (A+B)+C = A+(B+C).

How would I present a proof for this? Would this be good?:

If you present a proof, then you should always give the basic axioms you accept and the results that you've already proven. A lot of times it is clear from the context, but as somebody new to proofs, you should really give all the references. So: what axioms do you accept? What theorems do you accept? You refer later to something that is already proven, you should state explicitely the result you've proven and you should state it as a lemma before the proof.

So the proof should be like this:
AXIOMS
A list of the axioms, or a reference

LEMMAS
A list of all the lemmas you accept, or a reference

THEOREM
The theorem you want to prove
PROOF
The proof

As you advance in math, the axioms and lemmas will usually be clear from context, but here you should give them.

If (A + B) = C , then A = (C - B)

Why? You should justify this.

If (A - B) = (C - 2B) , then (A - B) + B = (C - 2B) + B (which gives us A = (C - B))

Your proof is backwards. Now you start by assuming that A-B = C-2B. You can't assume that, you need to prove it.
Let's say you need to show that "If it rains, then always walk with an umbrella".
You show this by starting off with: "I am walking with an umbrella, thus..."
This is clearly not correct. You can't start your proof with "If A-B=C-2B".
You need to start of with "If A+B=C" and then do some things and eventually end up with "and thus A-B=C-2B".

As previously demonstrated , if A = (C - B) , then (A + B) = C , therefore proving that if A , B and C are real numbers and that (A + B) = C , then (A - B) = (C - 2B).
I am just a newbie trying to learn so go easy on me.

 

[Gatsby88]

You should really start from some assumptions and work through and it seems as though youve assumed it to be true and worked backwards. Working backwards is good for proof by contradiction, but for a straightforward proof like this just go through the algebra.

For example, your starting points are that
1. A, B and C are real (this doesn't really affect anything in here, but there are other types of number where some of the basic rules we use in this proof don't work)
2. A + B = C

Given this, rearrange (2) to get A = C - B (as youve done) and B = C - A

Then A - B = (C - B) - (C - A) = (C - B) - (C - (C - B)) = C - 2B as required.

[Apologies someone got there whilst I was typing]

 

[reenmachine]

I greatly appreciate the feedbacks.I knew in the back of my mind that I also had to justify that if A+B=C , then A=C-B.

As for the ( ) , I did it so it would be clearer on the forum , I don't normally put them.I was unaware it was important not to put them in these circumstances.If A,B,C are real numbers and that A+B=C , I will attempt to prove that A-B is always equal to C-2B.

Why? You should justify this.

let me give it a try:

If A+B = C , then (A+B)-B = C-B = A.
Since C-B = A , then C-B-B = A-B = C-2B.
Thus , if A,B,C are real numbers and A+B=C , A-B will always be equal to C-2B.

Is this correct or I also have to justify something else?

 

[reenmachine]

You should really start from some assumptions and work through and it seems as though youve assumed it to be true and worked backwards. Working backwards is good for proof by contradiction, but for a straightforward proof like this just go through the algebra.

For example, your starting points are that
1. A, B and C are real (this doesn't really affect anything in here, but there are other types of number where some of the basic rules we use in this proof don't work)
2. A + B = C

Given this, rearrange (2) to get A = C - B (as youve done) and B = C - A

Then A - B = (C - B) - (C - A) = (C - B) - (C - (C - B)) = C - 2B as required.

[Apologies someone got there whilst I was typing]

Thanks a lot for the advices , very appreciated! I should've done the B = C-A part.

 

[micromass]

I greatly appreciate the feedbacks.I knew in the back of my mind that I also had to justify that if A+B=C , then A=C-B.

As for the ( ) , I did it so it would be clearer on the forum , I don't normally put them.I was unaware it was important not to put them in these circumstances.


If A,B,C are real numbers and that A+B=C , I will attempt to prove that A-B is always equal to C-2B.



let me give it a try:

If A+B = C , then (A+B)-B = C-B = A.
Since C-B = A , then C-B-B = A-B = C-2B.
Thus , if A,B,C are real numbers and A+B=C , A-B will always be equal to C-2B.

Is this correct or I also have to justify something else?

Is this even close to acceptable to prove that if A+B=C , then C-B=A? I have the feeling it's completely incomplete.

OK, but which axioms are you accepting as true?? I can't say anything unless I know what you're supposed to accept and what not.

 

[reenmachine]

OK, but which axioms are you accepting as true?? I can't say anything unless I know what you're supposed to accept and what not.

To be perfectly honest I had no clue what you were talking about.

I looked it up a bit and am still confused about what axioms really are and which one I should use.

Are you looking for something like this?:

Axioms that I accept as true: A=A , B=B , C=C , A+B=C ?

Or should I include C-B=A , C-A=B , B-C=-A and A-C=-B?

 

[micromass]

To be perfectly honest I had no clue what you were talking about.

I looked it up a bit and am still confused about what axioms really are and which one I should use.

Are you looking for something like this?:

Axioms that I accept as true: A=A , B=B , C=C , A+B=C ?

Or should I include C-B=A , C-A=B , B-C=-A and A-C=-B?

You can't prove anything without accepting some statements as true. When working with the real numbers, you usually take the field axioms as axioms. See http://mathworld.wolfram.com/FieldAxioms.html
These statements should be accepted as true. There are other axioms too, but those won't be needed here.

The idea is to justify every step by using either these axioms or proven results.

 

[WannabeNewton]

He is talking about the field axioms for $\mathbb{R}$ reenmachine. Here is a list of the field axioms as well as the order axioms and the least upper bound property: http://www.math.sjsu.edu/~hsu/courses/131a/axioms.pdf

 

[reenmachine]

I see , thank you all very much for your help! I'm still a long way from proving theorems :D

Another quick question , when I present the axioms that I accept as true , do I have to write that it's distributivity or I simply state that I believe that a(b+c) = ab+ac ? Or can I simply say that I accept the Field axioms for R as all true?

If I believe all the Field axioms to be true , do I still need to justify that if A+B=C then C-B=A? Or is the fact I mentioned I believe all the field axioms are true enough to justify it?

 

[micromass]

I see , thank you all very much for your help! I'm still a long way from proven theorems :D

Another quick question , when I present the axioms that I accept as true , do I have to write that it's distributivity or I simply state that I believe that a(b+c) = ab+ac ?

Both are acceptable. If you want to explain a step in a proof, then you should just write distributivity.

For example: "From distributivity follows that $2(x+3) = 2x + 6$.

 

[reenmachine]

Both are acceptable. If you want to explain a step in a proof, then you should just write distributivity.

For example: "From distributivity follows that $2(x+3) = 2x + 6$.

That's great! I really have to start learning the axioms more in depth.

any suggestions about what else should I look into? (like axioms)

 

[micromass]

That's great! I really have to start learning the axioms more in depth.

Any suggestions on how to approach self-teaching as far as writing mathematical proofs?

The best thing to do is to write as many proofs as you can. After you made a proof, you should show it to somebody on PF (or elsewhere) and let him comment (= rip it apart). You'll learn proofs pretty fast that way.

 

[reenmachine]

I finished high school around 9 or 10 years ago , and in my area high school is 5 years (following the 6 or 7 years in elementary school) therefore it ends in 11th or 12th grade but I'm not sure of the exact structure compared to the US.

Is it normal that I didn't learn to write proofs in high school back then? How ready to write proofs are the undergraduate math students (at the start of undergraduate)?

 

[Fredrik]

I think that when you're just getting started with these things, you should write everything out carefully, and only use one axiom or theorem in each step. For example, if you want to solve the equation 5+x=3, you write something like this:

Assumption: $5+x=3$

Every number has an additive inverse. The additive inverse of 5 is denoted by -5.

You can always add the same number to both sides: $-5+(5+x)=-5+3$

Computation of -5+3: $-5+(5+x)=-2$

Addition is associative: $(-5+5)+x=-2$

The defining property of additive inverses: $0+x=-2$

0 is the additive identity: $x=-2$​

Is it normal that I didn't learn to write proofs in high school back then?

Yes. It would be normal today too, I think.

How ready to write proofs are the undergraduate math students (at the start of undergraduate)?

Not at all. They are terrible at it. They make mistakes that seem absurd to those of us with more experience, like to not use the definitions of the terms and notations in the statement they're supposed to prove. That's probably the worst mistake that's really common. Another very common mistake is to not make it clear if their variables are part of "for all" statements, "there exists" statements, or if they simply have some specific value.

 

[micromass]

I finished high school around 9 or 10 years ago , and in my area high school is 5 years (following the 6 or 7 years in elementary school) therefore it ends in 11th or 12th grade but I'm not sure of the exact structure compared to the US.

Is it normal that I didn't learn to write proofs in high school back then? How ready to write proofs are the undergraduate math students (at the start of undergraduate)?

In the US, it's very normal not to write proofs in high school and most undergrads are very underprepared. So don't worry if you don't know any proofs.
In my country, proofs are done in high school, so undergrads have much more experience with them and are better prepare.

Proofs really aren't very difficult. You should just put some effort in it. You'll learn it faster than you think.

 

[Fredrik]

If I believe all the Field axioms to be true , do I still need to justify that if A+B=C then C-B=A? Or is the fact I mentioned I believe all the field axioms are true enough to justify it?

A minor nitpick. You don't have to believe them to be objectively true. The right way to think is that a set with an addition operation and a multiplication operation is called a field if it satisfies the axioms. Then you can prove that if A,B,C are members of a field, then A+B=C implies that C-B=A, while completely disregarding the issue of whether there really is such a thing as a "field".

This particular statement is proved the way I did it for the special case 5+x=3 above. Once you have done that for arbitrary A,B,C, the statement is considered a theorem. If you need to use it in another proof, you simply refer to this theorem instead of to the axioms.

 

[reenmachine]

thanks to both of you again!

My goal would be to capitalize on the months free of school that I have to try to learn as much as I can about proofs and math in general so hopefully I'll be ahead of the curve when school starts.I'm very confident about my mathematical abilities but I also greatly improved my work ethics which is what I was lacking when I was younger.

Please excuse my english , it is my second language.

 

[reenmachine]

A minor nitpick. You don't have to believe them to be objectively true. The right way to think is that a set with an addition operation and a multiplication operation is called a field if it satisfies the axioms. Then you can prove that if A,B,C are members of a field, then A+B=C implies that C-B=A, while completely disregarding the issue of whether there really is such a thing as a "field".

This particular statement is proved the way I did it for the special case 5+x=3 above. Once you have done that for arbitrary A,B,C, the statement is considered a theorem. If you need to use it in another proof, you simply refer to this theorem instead of to the axioms.

Not sure if I'm following you correctly.

When you're saying ''a set with an addition operation and a multiplication operation is called a field if it satisfies the axioms'' , does it need to include both addition and multiplication operations? How could an addition or multiplication operation not satisfy the axioms? Is there only one sort of field? If not , do you have to specify which kind of field it is? If A and B are members of a field , is it possible that C isn't if A+B=C?

I'm sorry for the countless stupid questions , I'm just trying to get the big picture and this is my way :X

 

[Fredrik]

Not sure if I'm following you correctly.

When you're saying ''a set with an addition operation and a multiplication operation is called a field if it satisfies the axioms'' , does it need to include both addition and multiplication operations? How could an addition or multiplication operation not satisfy the axioms? Is there only one sort of field? If not , do you have to specify which kind of field it is? If A and B are members of a field , is it possible that C isn't if A+B=C?

I'm sorry for the countless stupid questions , I'm just trying to get the big picture and this is my way :X

No problem, these are good questions. The definition of "field" can be stated like this:

An ordered triple (F,A,S) is said to be a field if F is a set, A and M are functions from F×F into F, and the following statements are true.

1. For all x,y,z in F, A(x,A(y,z))=A(A(x,y),z).
2. There's a z in F such that for all x in F, A(x,z)=A(z,x)=x.
...
and so on.

To simplify the notation, we write x+y instead of A(x,y) and xy instead of M(x,y). We call A "addition" and M "multiplication". The special element that's guaranteed to exist by axiom 2 is called the additive identity, or simply "zero", and is denoted by 0.

As an example of how a given $A:F\times F\to F$ might not satisfy the axioms, let's take F to be the set $\mathbb R^2$ that consists of all ordered pairs $(x_1,x_2)$ such that $x_1,x_2$ are real numbers, and define A by $A(x,y)=x_1$ Since $$A((1,0),(2,0))=1\neq 2=A((2,0),(1,0))$$ it's not true that A(x,y)=A(y,x) for all x,y in F.

There are many examples of triples that satisfy all the axioms, so there are many fields. If F and G are fields, and there exists a bijective function $\phi:F\to G$ such that $\phi(x+y)=\phi(x)+\phi(y)$ and $\phi(xy)=\phi(x)\phi(y)$ for all x,y in F, the two fields are said to be isomorphic. Two isomorphic fields can be thought of as "the same for all practical purposes".

There's also something called an "ordered field". This is a 4-tuple (F,+,·,<) such that the triple (F,+,·) is a field and < is a relation on F that satisfies a couple of additional axioms. The relation < can be used to define what it means for a set to be "bounded from above" and to have a "least upper bound". An ordered field F is said to be complete (or Dedekind complete) if every subset of F that's bounded from above has a least upper bound. It turns out that all complete ordered fields are isomorphic to each other. (The definition of "isomorphism" for ordered fields includes the requirement that $\phi(x)<\phi(y)$ for all x,y such that x<y).

Now we have two options: We can define the field of real numbers ℝ as a specific complete ordered field, or we can say that the members of any complete ordered field can be called "real numbers". I prefer the latter, but some people prefer the former.

I realize that I didn't include a lot of details at the end, but I have to leave.

 

[reenmachine]

No problem, these are good questions. The definition of "field" can be stated like this:

An ordered triple (F,A,S) is said to be a field if F is a set, A and M are functions from F×F into F, and the following statements are true.

1. For all x,y,z in F, A(x,A(y,z))=A(A(x,y),z).
2. There's a z in F such that for all x in F, A(x,z)=A(z,x)=x.
...
and so on.

To simplify the notation, we write x+y instead of A(x,y) and xy instead of M(x,y). We call A "addition" and M "multiplication". The special element that's guaranteed to exist by axiom 2 is called the additive identity, or simply "zero", and is denoted by 0.

As an example of how a given $A:F\times F\to F$ might not satisfy the axioms, let's take F to be the set $\mathbb R^2$ that consists of all ordered pairs $(x_1,x_2)$ such that $x_1,x_2$ are real numbers, and define A by $A(x,y)=x_1$ Since $$A((1,0),(2,0))=1\neq 2=A((2,0),(1,0))$$ it's not true that A(x,y)=A(y,x) for all x,y in F.

There are many examples of triples that satisfy all the axioms, so there are many fields. If F and G are fields, and there exists a bijective function $\phi:F\to G$ such that $\phi(x+y)=\phi(x)+\phi(y)$ and $\phi(xy)=\phi(x)\phi(y)$ for all x,y in F, the two fields are said to be isomorphic. Two isomorphic fields can be thought of as "the same for all practical purposes".

There's also something called an "ordered field". This is a 4-tuple (F,+,·,<) such that the triple (F,+,·) is a field and < is a relation on F that satisfies a couple of additional axioms. The relation < can be used to define what it means for a set to be "bounded from above" and to have a "least upper bound". An ordered field F is said to be complete (or Dedekind complete) if every subset of F that's bounded from above has a least upper bound. It turns out that all complete ordered fields are isomorphic to each other. (The definition of "isomorphism" for ordered fields includes the requirement that $\phi(x)<\phi(y)$ for all x,y such that x<y).

Now we have two options: We can define the field of real numbers ℝ as a specific complete ordered field, or we can say that the members of any complete ordered field can be called "real numbers". I prefer the latter, but some people prefer the former.

I realize that I didn't include a lot of details at the end, but I have to leave.

Hmm thanks a lot for the answers.I admit that's a lot to digest , there's some symbols and concept I simply do not know in your post.

An ordered triple (F,A,S) is said to be a field if F is a set, A and M are functions from F×F into F, and the following statements are true.

Why the S instead of M in the ordered triple? Is it sure that it'll be an ordered triple , and not many more potentially? I'm not sure I understand what ''if F is a set'' means.A set of what? A and M are functions , that's fine , could D and S also be other functions (division , substraction) ? In this ordered triple (field) you can't substract? When you're saying they (A,M) are functions from F X F into F , what does it mean? What does the X between the two Fs means and what do the three F means for that matter? The two first F are functions into F which is a field? Why the X if I'm right?

When you are saying:

A(x,A(y,z))=A(A(x,y),z)

It means X+(Y+Z) = (X+Y)+Z ?

There's a z in F such that for all x in F, A(x,z)=A(z,x)=x.

The special element that's guaranteed to exist by axiom 2 is called the additive identity, or simply "zero", and is denoted by 0.

So basically , because of this statement , 0 exist in this particular field? What if no such statement would be made? Would it still be a field , a field without the additive identity?

I have many more questions for your last paragraph , but I feel like all the questions I've already asked are enough for the moment , I'll wait for some more info from you and try to understand these concept first!

I'm grateful to you all for taking the time , I'm really learning some good things because of you guys , it is priceless.

 

[Fredrik]

Why the S instead of M in the ordered triple?

Sorry, that was a moment of temporary confusion on my part. I have used the notation S before, in a similar explanation for vector spaces. In that context I chose the symbol S because that operation is called "scalar multiplication". Here it's just "multiplication", so I chose M, but ended up writing an S.

Is it sure that it'll be an ordered triple , and not many more potentially?

I'm not sure I understand this question. (F,A,M) is by definition a triple.

I'm not sure I understand what ''if F is a set'' means.A set of what? A and M are functions , that's fine , could D and S also be other functions (division , substraction) ? In this ordered triple (field) you can't substract? When you're saying they (A,M) are functions from F X F into F , what does it mean? What does the X between the two Fs means and what do the three F means for that matter? The two first F are functions into F which is a field? Why the X if I'm right?

OK, I see now that my post was way over your head. It's totally my fault, since you said that you only have a high school education in the title.

To understand what I said, you need to study the basics of set theory. In particular, you need to understand the concepts "function" and "cartesian product". In my country (Sweden) this would be covered in the first year at the university.

I will try to explain some of it briefly here. The problem with definitions of terms and notations in mathematics is that you can always ask what the terms and notations used in the definition means. If you answer those questions with more definitions, it will raise more questions of the same sort. This means that it's impossible to define everything. Something must be left undefined. Mathematicians have found that it's sufficient to leave two things undefined: What a set is, and what it means for a set to be a member of a set. Everything else is defined in terms of sets.

A function should be thought of as a "rule" that associates exactly one member of a set (called the codomain) with each member of a set (called the domain). The notation $f:X\to Y$ means that f is a function with domain X and codomain Y. Such an f is said to be a function from X into Y. (Note that this isn't the definition of the term "function". It's just some good advice about how to think of functions. The definition isn't relevant here).

The × is not an x. It's the symbol that's sometimes used for multiplication. Here it denotes the Cartesian product of two sets. $A\times B$ is the set of ordered pairs $(a,b)$ such that $a$ is a member of A and $b$ is a member of B.

A function from F×F into F associates exactly one member of F with each ordered pair (x,y) such that x and y are members of F. So if A is such a function, and x and y are members of F, A(x,y) is a member of F. If A is the addition operation of a field, we will use the notation x+y instead of A(x,y).

Subtraction in a field is defined by $x-y=x+(-y)$ where $-y$ is the additive inverse of $y$. Similarly, division by a non-zero element is defined by $x/y=xy^{-1}$ where $y^{-1}$ is the multiplicative inverse of $y$. This is why the axioms don't mention subtraction and division explicitly. They can be defined in terms of addition and multiplication.

When you are saying:

It means X+(Y+Z) = (X+Y)+Z right?

Yes.

So basically , because of this statement , 0 exist in this particular field? What if no such statement would be made? Would it still be a field , a field without the additive identity?

It wouldn't be called a field. There are terms for triples (F,A,M) that only satisfy some of the field axioms, but I don't think there's a term for a "field" without an additive identity. I think it has simply turned out to be more useful to drop some of the axioms for multiplication than to drop some of the axioms for addition. A triple (F,A,M) such that A satisfies the same axioms as the addition operation of a field, and M is associative and distributive over addition, is called a ring. (I don't know why that word was chosen). (Associative means that (xy)z=x(yz) and distributive over addition means that x(y+z)=xy+xz and (x+y)z=xz+yz).

The branch of mathematics that deals with definitions of these things (groups, rings, fields, etc.) is called abstract algebra.

 

[reenmachine]

Sorry, that was a moment of temporary confusion on my part. I have used the notation S before, in a similar explanation for vector spaces. In that context I chose the symbol S because that operation is called "scalar multiplication". Here it's just "multiplication", so I chose M, but ended up writing an S.

Great , that's what I was suspecting!

I'm not sure I understand this question. (F,A,M) is by definition a triple.

OK but if there would be something else than A and M , like for example (F,A,M,P) , would it be called a quadruple? Or is a quadruple impossible? (take note that P doesn't mean anything , just an exemple to be clear).

OK, I see now that my post was way over your head. It's totally my fault, since you said that you only have a high school education in the title.

No problem , I prefer too much than too little :)

To understand what I said, you need to study the basics of set theory. In particular, you need to understand the concepts "function" and "cartesian product". In my country (Sweden) this would be covered in the first year at the university.

This is where self-teaching mathematics gets tricky.''Studying the basics of set theory'' might not be that hard , but knowing where to find the right stuff in the right order is where it gets complicated.I will follow your advice and try to understand the concepts of function and cartesian product.

I will try to explain some of it briefly here. The problem with definitions of terms and notations in mathematics is that you can always ask what the terms and notations used in the definition means. If you answer those questions with more definitions, it will raise more questions of the same sort. This means that it's impossible to define everything. Something must be left undefined. Mathematicians have found that it's sufficient to leave two things undefined: What a set is, and what it means for a set to be a member of a set. Everything else is defined in terms of sets.

I understand your philosophical argument about finding a level from which to stop asking why , but I find it pretty confusing that we wouldn't know what a set is or what a member of set is if we use them...

A function should be thought of as a "rule" that associates exactly one member of a set (called the codomain) with each member of a set (called the domain). The notation $f:X\to Y$ means that f is a function with domain X and codomain Y. Such an f is said to be a function from X into Y. (Note that this isn't the definition of the term "function". It's just some good advice about how to think of functions. The definition isn't relevant here).

I'm not sure I understand everything , but let me give it a try and correct me if I'm wrong:
The domain is each member of a set , is the codomain also a domain if it's part of the set? In your example , X represents every member of the set and the function turns all the members of the set into Y , which is only a part of the set , therefore all members of a set becomes like the co-domain?

The × is not an x. It's the symbol that's sometimes used for multiplication. Here it denotes the Cartesian product of two sets. $A\times B$ is the set of ordered pairs $(a,b)$ such that $a$ is a member of A and $b$ is a member of B.

Oh I knew that X means multiplying sometimes , wasn't aware it was this kind of X in this example.Won't lie , the A and B part of your paragraph sounds simple but I don't understand.What is a compared to A? Why is a part of A? I guess I'm stupid because I'm still not sure what a set is suppose to be or what the purpose of having a set.I think I'm lacking context here.

A function from F×F into F associates exactly one member of F with each ordered pair (x,y) such that x and y are members of F. So if A is such a function, and x and y are members of F, A(x,y) is a member of F. If A is the addition operation of a field, we will use the notation x+y instead of A(x,y).

So basically both F doesn't mean the same thing? FxF are functions and F is the field , so in your previous ordered triple you had A & M , which would be A x M into F? (F,A,M)

Another quick point about your last sentence , if A is addition we'll use + instead of A , so why the need to ever write A (x,y)?

Subtraction in a field is defined by $x-y=x+(-y)$ where $-y$ is the additive inverse of $y$. Similarly, division by a non-zero element is defined by $x/y=xy^{-1}$ where $y^{-1}$ is the multiplicative inverse of $y$. This is why the axioms don't mention subtraction and division explicitly. They can be defined in terms of addition and multiplication.

Make sense

It wouldn't be called a field. There are terms for triples (F,A,M) that only satisfy some of the field axioms, but I don't think there's a term for a "field" without an additive identity. I think it has simply turned out to be more useful to drop some of the axioms for multiplication than to drop some of the axioms for addition. A triple (F,A,M) such that A satisfies the same axioms as the addition operation of a field, and M is associative and distributive over addition, is called a ring. (I don't know why that word was chosen). (Associative means that (xy)z=x(yz) and distributive over addition means that x(y+z)=xy+xz and (x+y)z=xz+yz).

Not sure I understand what you mean when you're saying ''there are terms that only satisfy some of the field axioms''.Do you mean like in the example you gave me about the x1 x2 etc...? How do you drop axioms for multiplication?

The branch of mathematics that deals with definitions of these things (groups, rings, fields, etc.) is called abstract algebra.

I'll try to look into it , again if I'm capable of finding some stuff in the right order and clear enough.

Thanks a lot Fredrik you seem like a sharp guy!

 

[Fredrik]

OK but if there would be something else than A and M , like for example (F,A,M,P) , would it be called a quadruple?

Yes, I think "quadruple" would be appropriate. You can also call it a "4-tuple". The term for an ordered set $(x_1,\dots,x_n)$ with n members is "ordered n-tuple". But I think the it's kind of superfluous to include the word "ordered", since people who say "n-tuple" always mean "ordered n-tuple".

This is where self-teaching mathematics gets tricky.''Studying the basics of set theory'' might not be that hard , but knowing where to find the right stuff in the right order is where it gets complicated.I will follow your advice and try to understand the concepts of function and cartesian product.

I don't think you will need to read a book on set theory just yet. What you need is a brief summary of the main ideas that's no more than ten pages long. Unfortunately I don't know a good place to find such a summary.

In addition to functions and cartesian products, you will also need to understand subsets, unions, intersections, complements, differences and Venn diagrams. You also need to understand the symbols $\forall$ ("for all"), $\exists$ ("there exists"), $\in$ ("is a member of").

I'll explain a few things here. A set can be thought of as a collection of objects. There are two ways to specify a set. The first is to explicitly list its members. For example, the set whose members are 1, 2 and 3 is denoted by {1,2,3}. The second is to specify another set and a property that some of its members has. For example, let $\mathbb Z$ be the set of all integers. The set of all odd integers can be written as $\{n\in\mathbb Z|n\text{ is odd}\}$.

Two sets A and B are equal if and only if every member of A is a member of B and every member of B is a member of A. This means that {1,2,3}={3,1,2} and that {1,2,2,2,7}={7,7,2,1}={1,2,7}.

Ordered n-tuples are a lot like sets that are specified in the first of the two ways discussed above, but the condition for equality is different than the corresponding condition for sets. For example, the triples (a,b,c)=(d,e,f) are equal if and only if a=d, b=e and c=f. It's also impossible for an n-tuple to be equal to an m-tuple unless m=n.

I understand your philosophical argument about finding a level from which to stop asking why , but I find it pretty confusing that we wouldn't know what a set is or what a member of set is if we use them...

You'll get used to it.

Note that the alternative is to not know what a real number is, what an integer is, etc. In the context of set theory, these things can all be given exact definitions.

The domain is each member of a set , is the codomain also a domain if it's part of the set?

The domain is a set. The codomain of a function $f:X\to Y$ is also the domain of some function $g:Y\to Z$. The domain and codomain of a function f is often the same set. Consider e.g. the function $f:\mathbb R\to\mathbb R$ defined by f(x)=x+3 for all x in ℝ. Its domain is the set of all real numbers, and so is its codomain.

In your example , X represents every member of the set and the function turns all the members of the set into Y

X is a set. You can say that f associates exactly one member of Y with each member of X, or that f "takes" members of X to members of Y. If f(x)=x+3 for all x in ℝ, then f takes 3 to 7, $\pi$ to $\pi+3$, etc.

, which is only a part of the set , therefore all members of a set becomes like the co-domain?

I don't understand this part.

the A and B part of your paragraph sounds simple but I don't understand.What is a compared to A? Why is a part of A?

$a$ is a member of A. (Another way to say that is $a$ is an element of $A$). Why is it a member? Because I said so. Let's look at what I said again. I said that $A\times B$ is defined as the set of all $(a,b)$ such that $a\in A$ and $b\in B$. ($\in$ means "is a member of"). In the fancy notation,
$$A\times B=\{(a,b)|a\in A,\, b\in B\}.$$ (This notation actually violates the rules I just mentioned about how to specify a set, because I'm not mentioning what set the pairs (a,b) belong to, but since I know that there is such a set, it's safe to say that this abuse of notation won't cause any problems later). One of the most useful examples is the set
$$\mathbb R^2= \mathbb R\times\mathbb R =\{(x,y)|x,y\in\mathbb R\}.$$ The set $R$ can be interpreted as a line. The set $\mathbb R^2$ can be interpreted as a plane.

I'm still not sure what a set is suppose to be or what the purpose of having a set.I think I'm lacking context here.

Don't worry, you will encounter thousands of examples when you study math.

So basically both F doesn't mean the same thing? FxF are functions and F is the field , so in your previous ordered triple you had A & M , which would be A x M into F? (F,A,M)

No, F is a set, and F×F is a set. Specifically, it's the set of all ordered pairs (x,y) such that both x and y are members of F. A and M are the functions, both with domain F×F and codomain F. I indicated this by writing $A:F\times F\to F$ and $M:F\times F\to F$. It's the triple (F,A,M) that may or may not be a field, depending on whether it satisfies the field axioms or not. If (F,A,M) is a field, the proper way to refer to F is as "the underlying set of the field (F,A,M)", but no one actually does that. It's standard to abuse the terminology by referring to F as a field if (F,A,M) is a field. Similarly, when people talk about members "of the field", they mean members of its underlying set F.

Another quick point about your last sentence , if A is addition we'll use + instead of A , so why the need to ever write A (x,y)?

I just thought the explanation would be clearer that way. Note that you don't know if A is the addition operation of a field until after you have checked if it satisfies the field axioms. So you don't know at the start if the + notation is appropriate.

Not sure I understand what you mean when you're saying ''there are terms that only satisfy some of the field axioms''.Do you mean like in the example you gave me about the x1 x2 etc...? How do you drop axioms for multiplication?

No, I meant like in the definition of "ring" that I included in my previous post. (I didn't say exactly what you put in quotes. I didn't mention "terms" that may or may not satisfy the field axioms. I mentioned that there are terms (i.e. words) reserved for triples (F,A,M) that only satisfy some of the axioms). A triple (F,A,M) is said to be a field if it satisfies all 9 of the field axioms. A triple (F,A,M) is said to be a ring if it satisfies the first 7 of the field axioms. That's what I meant by "dropping" some of the axioms.

The set of integers is a good example of a ring. It has an addition operation and a multiplication operation that satisfy 8 of the field axioms. The axiom that isn't satisfied is the one that says that every member of the set except 0 has a multiplicative inverse. For example, 1/3 isn't an integer.

You probably shouldn't spend too much time on understanding abstract algebra right now. This is typical third-year stuff. But you should study a few pages of set theory as soon as possible.

 

[reenmachine]

Alright I'm going to wait before ''attacking'' your last post's material and just ask some questions in relation to my reading on basic set theory.As for the moment I'm trying to learn symbols and get comfortable with the jargon and basic systems.

Here I go:

Suppose 5 subsets of U as I've illustrated with my super paint skills (end of the post) with the 5 circles A,B,C,D,N.I will now try to make many true statements using symbols:

A∩B ∈ A
A∩B ∈ B
N∩A = ∅
U - ((A∪C)∪B) = (A'∪C')∪B'
N∩U = N
(A,B,C,D) ∈ N'
D⊂A
D∈A
D∉B
B-B = ∅ ?
D∩A= D
A⊄A
N∪U= U
A∩B∩C ∈ C (or A or B)

I still have problems understanding power sets correctly.In my reading source (now ignoring my picture of circles) , they tell me that if A=(a,b) then ∅∉A.How could you be a subset of A (∅) yet not be a member of A? (they mention that ∅ is always a subset of any set A)

And what is the powerset of U? Is U itself it's own self-containing powerset?

thanks , hope I've done decent progress in 2 days to understand set theory a bit more.

 

[micromass]

Alright I'm going to wait before ''attacking'' your last post's material and just ask some questions in relation to my reading on basic set theory.As for the moment I'm trying to learn symbols and get comfortable with the jargon and basic systems.

Here I go:

Suppose 5 subsets of U as I've illustrated with my super paint skills (end of the post) with the 5 circles A,B,C,D,N.I will now try to make many true statements using symbols:

A∩B ∈ A

Incorrect. You're saying that $A\cap B$ is an element of $A$. This is not true (in general).

For example, let Group A be the collection of all people with blond hair. Let Group B be the collection of all girls. A member of $A\cap B$ is then a blond girl. A member of $A$ is a person with blond hair. You are saying now that $A\cap B\in A$ which means that $A\cap B$ is a blond person. However, $A\cap B$ is a collection and can't be a person at all! So what you wrote makes no sense.

You should have written that $A\cap B\subseteq A$. That means that every member of $A\cap B$ is also a member of $A$. Thus it means that every blond girl is also a person with blond hair. This is true.

Do you understand what you did wrong?? You made the same mistake in a lot of your other statements. Do you see where? Can you correct it?

I still have problems understanding power sets correctly.In my reading source (now ignoring my picture of circles) , they tell me that if A=(a,b) then ∅∉A.How could you be a subset of A (∅) yet not be a member of A? (they mention that ∅ is always a subset of any set A)

The group of all blond girls is a subset of the group of all girls. But the group of all blond girls is not a girl!

 

[Fredrik]

The symbol $\in$ shouldn't be used in any of these examples. For example, the first one should be $A\cap B\subset A$. If I replace all the $\in$ with $\subset$, then one of these statements says that $A\not\subset A$. This is true if what you mean by $\subset$ is "is a proper subset of", but false if you mean "is a subset of".

The others are correct (after replacing $\in$ by $\subset$), as far as I can tell after only taking a quick glance at them.

 

[reenmachine]

Do you understand what you did wrong?? You made the same mistake in a lot of your other statements. Do you see where? Can you correct it?

Let me give it a try:

A∩B ⊆ A
A∩B ⊆ B
N∩A = ∅
U - ((A∪C)∪B) = (A'∪C')∪B'
N∩U = N
(A,B,C,D) ⊆ N'
D⊂A
D∈A(Wrong!)
D∉B
B-B = ∅ ?
D∩A= D
A=A
A⊄A
A⊆A
N∪U= U
A∩B∩C ⊂ C (or A or B)

Both ⊂ and ⊆ could be applied to all cases except A⊆A.

?

 

[micromass]

U - ((A∪C)∪B) = (A'∪C')∪B'

This seems wrong. You should look at it again.

(A,B,C,D) ⊆ N'

This is also wrong. You have a 4-tuple being a subset of N'. This is not valid. It is true that A is a subset of N' and the same for B,C and D. It is also true that

A\cup B\cup C\cup D\subseteq N^\prime

But what you wrote down is not correct.

A⊄A

You have to be careful here. Many people use $A\subset B$ to denote that A is a subset of B or equal to B. So in that case, $A\subset A$ is valid.
Other people (like me) hate that notation and use $A\subset B$ to denote that A is a subset of B but not equal to B. In that case, what you wrote down is correct.

What notation people use depends a lot on the specific author. It is usually clear from context though. But just be aware that $A\subset B$ can mean two different things depending on who writes it.

 

[reenmachine]

This seems wrong. You should look at it again.

Why is it wrong? If from the universal set you take out AuBuC , then what you have left is everything that wasn't A , B or C , therefore the union of A' , B' and C'.

This is also wrong. You have a 4-tuple being a subset of N'. This is not valid. It is true that A is a subset of N' and the same for B,C and D. It is also true that

A\cup B\cup C\cup D\subseteq N^\prime

But what you wrote down is not correct.

The union between multiple sets has to be guaranteed for these multiple sets to become a subset?

You have to be careful here. Many people use $A\subset B$ to denote that A is a subset of B or equal to B. So in that case, $A\subset A$ is valid.
Other people (like me) hate that notation and use $A\subset B$ to denote that A is a subset of B but not equal to B. In that case, what you wrote down is correct.

What notation people use depends a lot on the specific author. It is usually clear from context though. But just be aware that $A\subset B$ can mean two different things depending on who writes it.

Ok thanks! I will try to adjust depending on the context.

 

[Fredrik]

Why is it wrong? If from the universal set you take out AuBuC , then what you have left is everything that wasn't A , B or C , therefore the union of A' , B' and C'.

These rules are called de Morgan's laws: $(A\cap B)'=A'\cup B'$ and $(A\cup B)'=A'\cap B'$. Your result should be similar to that.

The union between multiple sets has to be guaranteed for these multiple sets to become a subset?

I don't understand the question.

 

[reenmachine]

1)These rules are called de Morgan's laws: $(A\cap B)'=A'\cup B'$ and $(A\cup B)'=A'\cap B'$. Your result should be similar to that. 2)I don't understand the question.

1) I don't understand the logic.

(A∩B)' should be everything within the universe of discourse (U) except the intersection of A and B (or anything within that intersection).

A'∪B' should be everything within the universe of discourse (U) that isn't A or B (or within A or B).

If both statements are correct , what about the part of A and the part of B that aren't intersecting with the other set (A or B) but that aren't in A' or B' (because a part of A or B)?

2) Basically , if you have two sets or two subsets , you have to ''unionize'' them in order to transform them into a subset if I want to write that A and B are a subset of N' (with N being a set that has no intersection with neither A or B).So I can't write (A,B) ⊆ N' but have to write A∪B ⊆ N'.Is that correct?

 

[Curious3141]

1) I don't understand the logic.

(A∩B)' should be everything within the universe of discourse (U) except the intersection of A and B (or anything within that intersection).

A'∪B' should be everything within the universe of discourse (U) that isn't A or B (or within A or B).

If both statements are correct , what about the part of A and the part of B that aren't intersecting with the other set (A or B) but that aren't in A' or B' (because a part of A or B)?

Draw a Venn diagram. A' includes the area of B that doesn't lie in A (i.e. excluding the intersection of A and B). Similarly, B' includes the area of A that doesn't lie in B. Both A' and B' include the area in U that's outside both A and B. The union operation includes all these areas (without double-counting the common ones).

 

[reenmachine]

Draw a Venn diagram. A' includes the area of B that doesn't lie in A (i.e. excluding the intersection of A and B). Similarly, B' includes the area of A that doesn't lie in B. Both A' and B' include the area in U that's outside both A and B. The union operation includes all these areas (without double-counting the common ones).

I currently have no idea what that is , but even without it I understand your point and am embarrassed that I didn't see it right away :X

I'll look for infos on Venn Diagrams tomorrow.

Thanks

 

[Curious3141]

I currently have no idea what that is , but even without it I understand your point and am embarrassed that I didn't see it right away :X

I'll look for infos on Venn Diagrams tomorrow.

Thanks

http://en.wikipedia.org/wiki/Venn_diagram

No need to feel "embarrassed" - we're all here to learn.

 

[reenmachine]

http://en.wikipedia.org/wiki/Venn_diagram

No need to feel "embarrassed" - we're all here to learn.

LOL I already made a similar diagram in this thread earlier.Just didn't know it was a ''Venn Diagram''.

I often feel embarrassed when I don't ''get'' something as quick as I would like , but these feelings are normally short-lived and my thirst to learn/know comes back to replace them pretty quickly. :)

cheers

 

[Fredrik]

A Venn diagram (yes, you already drew one) should be enough to convince you that $(A\cup B)'=A'\cap B'$. The proper way to prove this identity is based on the axiom that says that two sets X and Y are equal, if every member of X is a member of Y, and every member of Y is a member of X. The proof goes like this:

Let $x\in (A\cup B)'$ be arbitrary. Since $x\in(A\cup B)'$, we have $x\notin A\cup B$. This implies that $x\notin A$ and $x\notin B$ (because if these two statements aren't both true, x would be a member of $A\cup B$). This implies that $x\in A'$ and $x\in B'$. This implies that $x\in A'\cap B'$.

Now let $x\in A'\cap B'$ be arbitrary. Since $x\in A'\cap B'$, we have $x\in A'$ and $x\in B'$. This implies that $x\notin A$ and $x\notin B$. This implies that $x\notin A\cup B$ (because if x had been a member of $A\cup B$, then it would be a member of A or B). This implies that $x\in (A\cup B)'$.

 

[reenmachine]

A Venn diagram (yes, you already drew one) should be enough to convince you that $(A\cup B)'=A'\cap B'$. The proper way to prove this identity is based on the axiom that says that two sets X and Y are equal, if every member of X is a member of Y, and every member of Y is a member of X. The proof goes like this:

Let $x\in (A\cup B)'$ be arbitrary. Since $x\in(A\cup B)'$, we have $x\notin A\cup B$. This implies that $x\notin A$ and $x\notin B$ (because if these two statements aren't both true, x would be a member of $A\cup B$). This implies that $x\in A'$ and $x\in B'$. This implies that $x\in A'\cap B'$.

Now let $x\in A'\cap B'$ be arbitrary. Since $x\in A'\cap B'$, we have $x\in A'$ and $x\in B'$. This implies that $x\notin A$ and $x\notin B$. This implies that $x\notin A\cup B$ (because if x had been a member of $A\cup B$, then it would be a member of A or B). This implies that $x\in (A\cup B)'$.

Very clear thank you again!

Where does this axiom comes from per say though (I understand the logic behind it , but not from which list of multiple axioms does it came from)? Is it an official accepted and necessary axioms in all of set theory?

 

[Fredrik]

Very clear thank you again!

Where does this axiom comes from per say though (I understand the logic behind it , but not from which list of multiple axioms does it came from)? Is it an official accepted and necessary axioms in all of set theory?

There are several different set theories, each defined by a set of axioms about sets, and a set of axioms about how we can obtain new theorems from the ones we already have. The latter is referred to as a "proof theory". I am very far from an expert in these matters, but I think I understand some of the basic ideas at least. There seems to be a lot of flexibility in how exactly the proof theory is defined. Given a set of axioms about sets, there are lots of different sets of axioms about how to prove theorems that are equivalent in the sense that they will agree about what statements will be considered theorems. Because of this, the proof theory is rarely even mentioned, and you can pretty much think of the set theory as being defined by the axioms about sets.

There's one specific set of axioms called ZFC (Zermelo, Fraenkel, and the axiom of Choice) that's powerful enough to include all the mathematics that you are likely to ever find interesting. This includes all the mathematics of physics, and a lot more. The specific axiom I was referring to is called the axiom of extensionality. It's #1 on the Wikipedia page I linked to.

If you find this stuff interesting, there are several good books about set theory. Hrbacek and Jech is a good choice. Goldrei may be even easier, at least when it comes to the construction of the number systems, but it doesn't go as deep into the theory. However, you don't need to study axiomatic set theory right now. What you need is to understand when sets are equal, and a little about how to get new sets from the ones we already have, e.g. by taking unions, complements, etc. People refer to this approach as "naive set theory".

 

[Curious3141]

However, you don't need to study axiomatic set theory right now. What you need is to understand when sets are equal, and a little about how to get new sets from the ones we already have, e.g. by taking unions, complements, etc. People refer to this approach as "naive set theory".

Just to add to Fredrik's excellent post, it might be interesting to you (reenmachine) that set theory as usually introduced at the school level is "naive set theory". It introduces students to the concept intuitively, e.g. "let C be the set of cats", without bothering with axioms or rigor in general. The reason this is called "naive" is because it's rather easy to trip it up by creating logical paradoxes, e.g. Russell's Paradox. That's why naive set theory had to be abandoned in favour of a more sound theory, e.g. ZFC which had rather stricter rules about how sets can be defined and manipulated.

 

[reenmachine]

Thanks to both of you!

This thread took me to some interesting and unknown places.This is how I love to learn.I understand the need to understand naive set theory before going deeper but I still like to see a little bit of what's ahead of me to motivate me to learn the basics in order to ''get there''.I will try to buy those book suggestions you gave me.

I will surely have more questions about these things , but first I'll try to finish the set theory basics read I pmed you (Fredrik).

cheers!

 

[WannabeNewton]

The cool thing is that with some mathematical maturity and a handle on proofs you could easily just start studying set theory since it has no other pre-requisites. Set theory is easily, in my opinion, the most beautiful of the mathematical branches. It is full of insanely elegant proofs (Cantor being the pervasive one in that department) and counter intuitive concepts (ordinals for example). The bible of all set theory books would have to be the one by Jech; it is very comprehensive but probably more than what you are looking for since it goes straight into ZFC. Good luck!

 

[reenmachine]

The cool thing is that with some mathematical maturity and a handle on proofs you could easily just start studying set theory since it has no other pre-requisites. Set theory is easily, in my opinion, the most beautiful of the mathematical branches. It is full of insanely elegant proofs (Cantor being the pervasive one in that department) and counter intuitive concepts (ordinals for example). The bible of all set theory books would have to be the one by Jech; it is very comprehensive but probably more than what you are looking for since it goes straight into ZFC. Good luck!

Thank you!

I have to face reality , I'm not getting any younger (mathematically speaking at least).I have no choice but to study by myself before returning to school , and even when I'm going to go to school hopefully I will be capable of learning more than what the programs has to offer in my free times.

At 26 years old with no university education , I can't simply wait here and pretend I'll rock everything on my way just by returning to school and doing like 95% of mathematics students.I have to breathe math , and for the moment my lungs are loving it.

Set theory strangely ressembles what logicians would work on.

edit: btw , when you're saying no pre-requisites are required to do set theory , does it mean set theory is a bit of an outsider as far as mathematic branches? Meaning you don't have to study for years and years before being able to understand the basics like some other branches might require?

I really enjoy the pure logic aspect of what I'm seeing right now.Is set theory a popular branch at the moment?

 

[WannabeNewton]

Set theory strangely ressembles what logicians would work on.

Axiomatic set theory does have a good chunk of formal logic. Most mathematics sequences I've seen usually start with a formal logic course before following up with an axiomatic set theory course.

 

[WannabeNewton]

edit: btw , when you're saying no pre-requisites are required to do set theory , does it mean set theory is a bit of an outsider as far as mathematic branches? Meaning you don't have to study for years and years before being able to understand the basics like some other branches might require?

I really enjoy the pure logic aspect of what I'm seeing right now.Is set theory a popular branch at the moment?

Well by no pre-requisites (other than formal logic depending on how deep you want to go) I meant you don't for example need to have done analysis or algebra beforehand to learn set theory. Set theory is a bit removed from the other branches if you get really deep. I don't know what your definition of "basics" are but to learn anything properly requires a good amount of time. As far as "popularity" I don't know anything about that. Micromass could probably answer that.

 

[reenmachine]

I just realized micromass is an high school student , he must be a sick mathematical talent.Great for him! If he sees this post maybe he can give me his take on the branch of set theory as far as popularity goes.

As for how deep I want to go , I guess it depends on what branch of mathematic I will ultimately choose , but even with my age I still dream of going as deep as I can in pure math.

 

[WannabeNewton]

I just realized micromass is an high school student , he must be a sick mathematical talent.Great for him!

Yes he is quite the brilliant high schooler

 

[reenmachine]

Yes he is quite the brilliant high schooler

I don't understand , why didn't they push him toward university?The guy looks like he is almost ready or ready to do serious research , but maybe that's just my own ignorance of what level you have to reach to do research :X

In any case I'm impressed.

 

[WannabeNewton]

I don't understand , why didn't they push him toward university?

He probably just wanted to enjoy his high school years while it lasted.

 

[reenmachine]

There's one specific set of axioms called ZFC (Zermelo, Fraenkel, and the axiom of Choice) that's powerful enough to include all the mathematics that you are likely to ever find interesting. This includes all the mathematics of physics, and a lot more. The specific axiom I was referring to is called the axiom of extensionality. It's #1 on the Wikipedia page I linked to.

If you find this stuff interesting, there are several good books about set theory. Hrbacek and Jech is a good choice. Goldrei may be even easier, at least when it comes to the construction of the number systems, but it doesn't go as deep into the theory.

Can you please suggest me some specific titles from these authors you think I should buy right now?

Also , do you happen to know a good book on formal logic that I should buy to get started with the basic concepts of formal logic?

(those questions applies for everybody who have suggestions)