Proving the Inequality 1 > 0 in Real Numbers

[issacnewton]

Hi

I am trying to prove that 1 > 0. I am going to assume a contradiction .

Assume $$1 \leqslant 0$$

First consider 1=0, Let $a \in \mathbb{R}$ be arbitrary. So

$$a=1.a = 0.a = 0$$ using field axioms. Actually 0.a=0 is not field axiom , but I have
proved it separately .

that means $$\forall \;\; a\in \mathbb{R} \Rightarrow a=0$$

since all numbers are equal to 0 , no inverse exists in the set R. So R is not a field. Hence
the contradiction. So

$$1 \neq 0$$

that means we have to consider the second option

$$1 < 0$$

Consider $a > 0 \backepsilon a \in R$

Now I have already proved another theorem.

If q,b,c are in R and q < b , c < 0 then qc > bc

letting c =1 and q = 0 , b= a we can say that

(0)(1) > (a)(1) which means 0 > a or a < 0 but this is contradiction since
we assumed that a> 0 . So our assumption that 1 < 0 , is wrong.

Hence 1 > 0

is my proof right ?

 

[micromass]

Hi IssacNewton!

It might help if you list the axioms you use. Some of the comments I make will be about how I've seen the field axioms, but you might have different axioms.

Hi

I am trying to prove that 1 > 0. I am going to assume a contradiction .

Assume $$1 \leqslant 0$$

First consider 1=0, Let $a \in \mathbb{R}$ be arbitrary. So

$$a=1.a = 0.a = 0$$ using field axioms. Actually 0.a=0 is not field axiom , but I have
proved it separately .

that means $$\forall \;\; a\in \mathbb{R} \Rightarrow a=0$$

since all numbers are equal to 0 , no inverse exists in the set R. So R is not a field. Hence
the contradiction.

This is not a contradiction. The field axioms say that every nonzero element in R must have an inverse. But taking R={0} satisfies this: every nonzero element has an inverse, because there are no nonzero elements!

In fact, most field axioms take $1\neq 0$ as an axiom. So there's nothing to prove here. However, your axioms might do things differently...

that means we have to consider the second option

$$1 < 0$$

Consider $a > 0 \backepsilon a \in R$

Who says such an a exists?

Now I have already proved another theorem.

If q,b,c are in R and q < b , c < 0 then qc > bc

letting c =1 and q = 0 , b= a we can say that

(0)(1) > (a)(1) which means 0 > a or a < 0 but this is contradiction since
we assumed that a> 0 . So our assumption that 1 < 0 , is wrong.

Hence 1 > 0

is my proof right ?

The rest of the proof looks right.

 

[issacnewton]

Hi Micro

Your comments about the first part are right. Only non zero elements of R have inverse.
I will try to modify that part. The axioms I am using doesn't say that $1 \neq 0$
as part of the any axiom. So I will need to come up with some contradiction.

For the second part, I am using some $a \in \mathbb{R}$ which is greater than 0. Is that wrong ?

 

[micromass]

Hi Micro

Your comments about the first part are right. Only non zero elements of R have inverse.
I will try to modify that part. The axioms I am using doesn't say that $1 \neq 0$
as part of the any axiom. So I will need to come up with some contradiction.

Your axioms somehow need to rule out the possibility that R={0}. So try to see why this can't be possible.

For the second part, I am using some $a \in \mathbb{R}$ which is greater than 0. Is that wrong ?

That isn't wrong, and I'm being extremely pedantic here. But who says that there exists an a which is greater than 0? That is, can't it happen that all elements in $\mathbb{R}$ are <0?? (it can't, but why not?)

 

[issacnewton]

Here's another attempt for the first part

Let $a,b \in \mathbb{R}$ and $a < b$

then $$1.a < 1.b$$ since 1 is multiplicative identity

$$0.a < 0.b$$ since 1=0

$$0 < 0$$ a.0 = 0 (I proved this elsewhere)

this is a contradiction. Hence

$$1 \neq 0$$

does it look right now ?

 

[micromass]

Here's another attempt for the first part

Let $a,b \in \mathbb{R}$ and $a < b$

Hmm, you're going to find me annoying. But how do you know there exists numbers in $\mathbb{R}$ such that a<b?

May I know what field axioms you're starting from?

 

[issacnewton]

That isn't wrong, and I'm being extremely pedantic here. But who says that there exists an a which is greater than 0? That is, can't it happen that all elements in $\mathbb{R}$ are <0?? (it can't, but why not?)

All elements of $\mathbb{R}$ can't be < 0 since we need an identity element for the operation of addition. One of the axioms says that there exists a zero element(
identity element) for the operation of addition.

 

[micromass]

All elements of $\mathbb{R}$ can't be < 0 since we need an identity element for the operation of addition. One of the axioms says that there exists a zero element(
identity element) for the operation of addition.

OK, but a=0 won't be good in your proof. You'll need to find an element >0...

 

[issacnewton]

here are the axioms I am using
1) for all $a,b \in \mathbb{R}$ we have $a+b , a.b \in \mathbb{R}$

2)$$\forall a,b \in \mathbb{R} , a+b=b+a \;\; a.b=b.a$$

3)$$\forall a,b,c \in \mathbb{R} ,\; (a+b)+c=a+(b+c) \;\; (a.b).c=a.(b.c)$$

4)there exists a zero element in $\mathbb{R}$ , denoted by 0, such that
$a+0=a \;\; \forall a \in \mathbb{R}$

5)$\forall a \in \mathbb{R}$ there exists an element -a in
$\mathbb{R}$ , such that $a+(-a)=0$

6) there exists an element in $\mathbb{R}$ , which we denote by 1, such that
$a.1=a \;\; \forall a\in \mathbb{R}$

7)$\forall a \in \mathbb{R}$ with $a\neq 0$ there exists an element
in $\mathbb{R}$ denoted by $\frac{1}{a}$ or $a^{-1}$
such that

$$a.a^{-1}=1$$

8)$\forall a,b,c \in \mathbb{R}$ we have

$$a.(b+c)=(a.b)+(a.c)$$

 

[Hurkyl]

here are the axioms I am using

Okay. Then your first challenge is to prove the following statement is not true:

$$\forall a,b \in \mathbb{R}: a = b$$​

(hint: this challenge is actually impossible to complete)

(also, those axioms don't say anything about > anyways, so you couldn't possibly use them to prove anything about orderings)

 

[micromass]

I'm afraid that this list of axioms is incomplete then, since R={0} satisfies all these axioms. There should at least by a remark that says that 1 is distinct from 0, or even something that says that R has more than one element.

If you check wiki's entry for a field you'll see one of the following axioms:

Additive and multiplicative identity
There exists an element of F, called the additive identity element and denoted by 0, such that for all a in F, a + 0 = a. Likewise, there is an element, called the multiplicative identity element and denoted by 1, such that for all a in F, a · 1 = a. To exclude the trivial ring, the additive identity and the multiplicative identity are required to be distinct.

This last sentence is essential.

 

[Unit]

Sorry to interrupt, but strictly speaking you also need the order axioms:

1) Trichotomy property: for all a, b in R, only one of the following holds: a < b, b < a, or a = b.

2) Transitive property: for all a, b, c in R, if a < b and b < c, then a < c.

3) Additive property: for all a, b, c in R, if a < b, then a + c < b + c.

4) Multiplicative property: for all a, b, c in R, if a < b and 0 < c then ac < bc; if a < b and c < 0 then bc < ac.

EDIT: As Hurkyl and micromass have pointed out, the additive and multiplicative identities must be distinct in any field; 0 ≠ 1.

 

[issacnewton]

Hi Hurkyl

I think you are also saying that $\mathbb{R}\neq \{0\}$ . My starting point that 1=0 lead me to conclude that $\mathbb{R}= \{0\}$ . Now you guys are saying that this can't be true. I don't see which of the axioms are violated. Or do I have
to think about the completeness axiom ?

Also some comments about the latex thing. The latex output here doesn't look so nice. Is
there some maintenance going on ?

 

[issacnewton]

Unit, oh yeah,

I forgot about the order axioms. They are there.

 

[issacnewton]

I think the axiom no 7 listed by implicitly says that $1\neq 0$

Since we have $1.1^{-1}=1$ that means , inverse of 1 exists and so
it must be true that $1\neq 0$ . am I right ?

 

[Unit]

How do you know 1.1 is in R?

The only numbers you know are in R are 1, 0, and the additive inverse of 1. Try cases, noting that 1 times 1 is 1.

 

[micromass]

Hi Hurkyl

I think you are also saying that $\mathbb{R}\neq \{0\}$ . My starting point that 1=0 lead me to conclude that $\mathbb{R}= \{0\}$ . Now you guys are saying that this can't be true. I don't see which of the axioms are violated.

None of the axioms are violated, that's the point. There's an axiom missing that says that $1\neq 0$.

 

[issacnewton]

How do you know 1.1 is in R?

The only numbers you know are in R are 1, 0, and the additive inverse of 1. Try cases, noting that 1 times 1 is 1.

closure property. Since $1 \in \mathbb{R}$

$$1.1 \in \mathbb{R}$$

 

[Fredrik]

I think the axiom no 7 listed by implicitly says that $1\neq 0$

It doesn't. However, you proved in post #1 that if 1=0, the field is the trivial field {0}.

Most authors include the axiom 1≠0. Some don't. I would choose not to include it, and instead do what you did: Prove that the only field that doesn't satisfy this condition is trivial.

 

[issacnewton]

None of the axioms are violated, that's the point. There's an axiom missing that says that $1\neq 0$.

Micromass, as I pointed out , doesn't that follow implicitly from axiom 7. go 2-3 posts back.

thanks

 

[Unit]

closure property. Since $1 \in \mathbb{R}$

$$1.1 \in \mathbb{R}$$

My mistake. You should use $1\cdot 1$ (that is, \cdot) to indicate multiplication. I mistook it 1.1 for 11/10.

 

[micromass]

Micromass, as I pointed out , doesn't that follow implicitly from axiom 7. go 2-3 posts back.

thanks

No, it doesn't follow from any of your axioms. Setting R={0} will satisfy all your axioms, so you explicitly need to eliminate the possibility that R is the trivial ring.

 

[issacnewton]

Micromass, so if I include $1\neq 0$ as one of the axioms, that takes care
of the first part of my proof. Which means that the statement $1=0$
must be wrong.

I remember reading axioms from other books and people do say explicitly that 1 is not 0.
So the book I am using has problems. Maybe I should write the author, Witold Kosmala

 

[micromass]

Micromass, so if I include $1\neq 0$ as one of the axioms, that takes care
of the first part of my proof. Which means that the statement $1=0$
must be wrong.

I remember reading axioms from other books and people do say explicitly that 1 is not 0.
So the book I am using has problems. Maybe I should write the author, Witold Kosmala

Hmm, I don't seem to find the book, but I guess it's a mistake. Writing him sounds like a nice thing to do!

 

[issacnewton]

Here's the link to the book

https://www.amazon.com/dp/0130457965/?tag=pfamazon01-20


and here's author webpage.

http://mathsci2.appstate.edu/~wak/

 

[Hurkyl]

I think the axiom no 7 listed by implicitly says that $1\neq 0$

First off, axiom 7 says every non-zero element has an inverse. If every element is zero, the axiom tells us nothing.

Secondly, in the zero ring, 0 does have an inverse: $0^{-1} = 0$.

(edit: didn't notice the second page)

 

[issacnewton]

Ok sorry for the late reply. I was traveling. We have already established that $1\neq 0$
Now assume the second option that $1 < 0$

$$\Rightarrow -1 > 0$$

Now I have proved the following theorem

Theorem: If $a \in \mathbb{R}$ such that $a \neq 0$ then
$a^2 > 0$.

I am going to use this theorem here. Since $-1 > 0$ it follows that
$(-1)^2 > 0$ hence $1 > 0$ which is a contradiction. Hence
we reject $1 < 0$ hence $1> 0$

is it right ?

 

[micromass]

Ok sorry for the late reply. I was traveling. We have already established that $1\neq 0$
Now assume the second option that $1 < 0$

$$\Rightarrow -1 > 0$$

Now I have proved the following theorem

Theorem: If $a \in \mathbb{R}$ such that $a \neq 0$ then
$a^2 > 0$.

I am going to use this theorem here. Since $-1 > 0$ it follows that
$(-1)^2 > 0$ hence $1 > 0$ which is a contradiction. Hence
we reject $1 < 0$ hence $1> 0$

is it right ?

Sounds good! Maybe you should show that $(-1)^2=1$...

 

[Unit]

Maybe you should show that $(-1)^2=1$...

My mistake; I misread your post.

OP: Well done.