Question about proof from a guy with a highschool education

1. Mar 31, 2013

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.

Last edited: Mar 31, 2013
2. Mar 31, 2013

micromass

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

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).

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.

Why? You should justify this.

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".

3. Mar 31, 2013

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 doesnt really affect anything in here, but there are other types of number where some of the basic rules we use in this proof dont 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]

4. Mar 31, 2013

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.

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?

Last edited: Mar 31, 2013
5. Mar 31, 2013

reenmachine

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

6. Mar 31, 2013

micromass

Staff Emeritus
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.

7. Mar 31, 2013

reenmachine

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?

8. Mar 31, 2013

micromass

Staff Emeritus
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.

9. Mar 31, 2013

WannabeNewton

10. Mar 31, 2013

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?

Last edited: Mar 31, 2013
11. Mar 31, 2013

micromass

Staff Emeritus
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 $2(x+3) = 2x + 6$.

12. Mar 31, 2013

reenmachine

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)

13. Mar 31, 2013

micromass

Staff Emeritus
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.

14. Mar 31, 2013

reenmachine

I finished highschool around 9 or 10 years ago , and in my area highschool 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 highschool back then? How ready to write proofs are the undergraduate math students (at the start of undergraduate)?

15. Mar 31, 2013

Fredrik

Staff Emeritus
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: $x=-2$​

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

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.

16. Mar 31, 2013

micromass

Staff Emeritus
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.

17. Mar 31, 2013

Fredrik

Staff Emeritus
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.

18. Mar 31, 2013

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.

19. Mar 31, 2013

reenmachine

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

20. Mar 31, 2013

Fredrik

Staff Emeritus
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.