Why is 1 greater than 0? Show the proof.

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Solution 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.

Solution 2

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 $1 = 1^2 > 0$!

Solution 3

It follows directly from Peano's axioms.

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

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Updated on July 19, 2020

Comments

  • Jane
    Jane over 3 years

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

    • Davis Yoshida
      Davis Yoshida about 7 years
      A proof depends on which axioms you need to use. You could take $1 > 0$ as an axiom and declare victory but that probably isn't what is required.
    • Jane
      Jane about 7 years
      well, so 1 > 0 is not an axiom and i have to prove it using axioms.. and i dont have idea how to prove it..
    • Jane
      Jane about 7 years
      well, we use basic axioms...
    • Arunas
      Arunas about 7 years
      This really is tricky without knowing which axioms you would be using - that would likely be in a reference text that you are using. If you look at the Peano Axioms ( en.wikipedia.org/wiki/Peano_axioms#Inequalities ), you might find something that helps.
    • GEdgar
      GEdgar about 7 years
      Questions like this are best asked of the instructor of the course. He knows what axioms are used in that course. We do not. Different treatments may start with different axioms. But (we hope) they all arrive at the same place eventually.
  • Jane
    Jane about 7 years
    Thanks. I found something like that:
  • Jane
    Jane about 7 years
    6 down vote 1=121=12 and x2>0x2>0 if x≠0x≠0. For the last claim, the main point is that by definition of positive x,y>0x,y>0 implies xy>0xy>0. This is used as follows: If x>0x>0 then x2=x⋅x>0x2=x⋅x>0. If x<0x<0 then −x>0−x>0 and so x2=(−x)2>0x2=(−x)2>0. In turn, this follows from (−x)y=−(xy)(−x)y=−(xy), which follows from distributivity: 0=(−x+x)y=(−x)y+(xy)0=(−x+x)y=(−x)y+(xy).
  • Jane
    Jane about 7 years
  • Jane
    Jane about 7 years
    well, i dont have any expierence with that, so..
  • Jane
    Jane about 7 years
    We have to prove that in R: