2
$\begingroup$

Prove that this is no real number such that $x \leq a$ for all real $x$.

I want to know if the way I proved it is valid or not.

Proof. We first prove that there is no real number $a$ such that $x=a$ for all real $x$. If this were true, then all real numbers would be equal to $a$. This is not possible, because the axiom that garantees the existence of identity elements tells us that the set of real numbers has the numbers $0$ and $1$. Therefore, the set of real numbers has more than only one element.

If $x$x$, then, of course $a>0$. It can be proven that $1>0$. There is an axiom that tells that if two real numbers are in $R^{+}$, the sum of the two is in $R^{+}$. So, $a+1>0$ is in in $R^{+}$ and in $R$. But we assumed that $a$ is greater than all the real numbers. Thus, $a>a+1$, which is an absurd.

Is this proof correct? Can I improve it? Thank you.

  • 4
    Why not just say that if such $a$ exist then since $a+1$ is also a real number (by axiom, since both $a$ and $1$ are real) and if $a+1\leq a$ it would imply $1\leq 0$ which you said you can prove otherwise. (the first part is not relavent as far as I can tell)2012-08-08
  • 8
    Your proof is no good at all. "$x\le a$ for all $x$" means "$(x < a \hbox{ or } x = a)$ for all $x$". This is not the same as "$(x for all $x)$ or $(x=a$ for all $x)$", as you seem to think.2012-08-08
  • 0
    @MJD You are of course correct, but this makes little change to the proof. OP needs to change "more than only one element" to "at least one element different from a", and "for all real x" to "for all real x different from a".2014-07-07
  • 0
    I think you are missing a step. Why is a>a+1 absurd? This should be proven since it's the heart of your proof. How to prove it depends on what axioms you can use.2014-07-07

3 Answers 3