1. ## hello!!!

what is wrong with the following proof that 1 is the largest real number?
1 is the largest real number :
proof:
let x be the largest real number (want to prove that x=1) then $x\geq 1$
(since $1\in R$ and x is the largest)
so that ${x^2} \geq x$
hence, either ${x^2} > x or {x^2}=x$
but ${x^2} > x$ is impossible so ${x^2} = x$ and therefore
${x^2} - x=0$ or x(x-1)=0 so that x=0 or x=1 since 1>0 then x=1 and thus 1 is the largest real number

2. Why is it "impossible" that $x^2 > x$?

3. Originally Posted by Matt Westwood
Why is it "impossible" that $x^2 > x$?
Because he assumed $x$ as the largest number, hence $x^2$ cannot be larger than $x$.

Edit: I think it is wrong to assume that there is a largest element in [tex]\mathbb{R}[\math].

4. Oh yes, I see.

Not sure then. All that tells me is that *if* there is a largest real number, *then* it is no greater than 1. As we know there are real numbers greater than 1, it follows that there is no greatest real number. (That is, if you think you've got the greatest one, you can always find one that's even greater, i.e. by squaring it.)

Interesting little question. Nice one.

5. So the flaw is in the first line, when he suppose that there's a largest element in $\mathbb{R}$. Because we know that $\mathbb{R}$ is a field. $x+x$ should lie in $\mathbb{R}$ and if $x$ is the largest element then $x+x \leq x \Leftrightarrow x \leq 0$. Taking any number $c >0$ shows that assuming the existence of a greater element in $\mathbb{R}$ is false.
Edit : Alternatively, to end a bit more formally, take any $d. Clearly $d<0$. Also $d \cdot d \in \mathbb{R}$, but $d\cdot d >0$, which is impossible since $x$ is the greatest element and $x \leq 0$. Hence we can't suppose that there exists a greatest element in $\mathbb{R}$.