Why is 1 greater than 0? Show the proof.

Question

I would like to prove that 1 > 0. And I need to use axioms, could somebody help me? Thanks'

Answer 1

The usual proof relies on $a^2> 0$ for any nonzero real number $a$. If you can prove that as a lemma from the basic axioms, and you note that 1 is the multiplicative identity, you will be done.

Answer 2

It follows directly from Peano's axioms.

You might want to refer to the different axioms you use.

Answer 3

It does very much depends on what "basic axioms" you use. Believe it or not not all mathematicians use the same. YOu should list which ones you have.

From an algebra perspective, we can use the ordered field axioms.

There's a lot of things to prove.

Intermediate and nescessary things to prove but our big goal is:

a) For all $a\ne 0$ we know $a^2 > 0$.

Then we are done, as $1 = (1)^2 > 0$.

To prove a) we usually have an axioms $a > b; and x > 0$ then $ax > bx$. We also have if $a > b$ then $a + x > b+x$ for all x$.

From there we can prove $x > 0 \iff 0 > -x$ by noting $x > 0 \implies x-x > 0 -x$ so $0 > -x$.

We need to prove tedious little things such as $(-x)(y) = x(-y) = -xy$ but using the axiom $a(b+c) = ab + bc$ and that for all $a$ there is a unique $-a$ so that $a+(-a) = 0$. Then we need to prove $(-x)(-y) = xy$.

This is all to prove that if $a > b$ and $x < 0$ then $ax < ab$ (because $x < 0 \implies -x > 0 \implies a(-x) > b(-x) \implies -ax > -bx \implies -ax +ax + bx > -bx +ax +bx \implies $bx > ax$).

So with all that in mind we can prove: If $x > 0$ then $x^2 =x*x > 0*x = 0$. And if $x < 0$ then $x*x > 0*x$ so $x^2 > 0$. So as long as $x \ne 0$ we have $x^2 > 0$!

So $1 = 1^2 > 0$!

Answer 4

This is kinda tricky to answer without knowing the context in which you are trying to prove it in. If you are talking strictly purely, as in I have a number 0 and the number 1 (as we all know them), then the proof is from the Peano axioms:

Axiom 1: 0 is in $\mathbb{N}$

Axiom 2: If x is in $\mathbb{N}$ then x+1 is in $\mathbb{N}$

Axiom 3: Any subset of $\mathbb{N}$ that satisifies the above actually coincides with $\mathbb{N}$.

Axiom 4: If x in $\mathbb{N}$ then x+1 $\neq$ 0

Axiom 5: If x+1=y+1 then x=y.

These are basic axioms of the natural numbers, the assertion comes from the fact that since 1 is in $\mathbb{N}$, and by axiom 4 no successor of 1 is 0, 1 must be a successor of 0 and hence bigger than. (This is hand wavey, a sucessor of x is a number which comes after x, i.e x'=x+1, x''=(x+1)+1... and a true proof would go into depths about orderings and the word 'bigger')

tl;dr By axioms, any natural number can be obtained by starting with 0 and adding 1 to it a certain amount of times so 1 can't be less than 0 and it certainly isn't equal, ergo it is greater than.

I've a casual interest in set theory which is why my mind went this way, however for a thorough proof you'd have to start from the bottom and define everything and prove existence of things.

Answer 5

Suppose you are using the axioms for an ordered field, which is a field with a linear order that interacts with the field's arithmetic according to :

Rule 1. If $a<b$ then $a+c<b+c.$

Rule 2. If $a<b$ and $0<c$ then $ac<bc.$

Now it is axiom of fields that the additive identity $0$ is not equal to the multiplicative identity $1.$ So in an ordered field we must have $0<1$ or $1<0$ . To show that the case $1<0$ is untenable, we have :$$1<0\implies 0<-1$$ because $0=1+(-1)<0+(-1)=-1$ by Rule 1 with $a=1, b=0, c=-1.$ But then, using Rule 2, with $a=1, b=0, c=-1$ we have $$1<0 \implies (1<0 \text { and } 0<-1)\implies -1=1(-1)<0(-1)=0\implies -1<0.$$ So $1<0$ implies both $0<-1$ and $-1<0 , $ which is untenable for a linear order $<.$