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I think it has something to do with induction, but since they aren't natural numbers, I'm not sure how to go about proving it. It seems totally intuitive, but I honestly have no clue where to start

[edit] Full proof using @ThomasGrubb 's advice:

Since $r \gt x$, $(\exists y \in {R})$ s.t. $r = x + y$

$r^2 \gt x^2$ can then be re-written as $(x+y)^2 \gt x^2$ which is equivilent to

$x^2 + xy + y \gt x^2$

Since $y$ must be a positive number, $r^2 \gt x^2$

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  • $\begingroup$ Why do you think it has something to do with induction? $\endgroup$ – Henrik supports the community Feb 26 '18 at 21:39
  • $\begingroup$ Take your "$x<y$" and multiply by $x$. Take your "$x<y$" and multiply by $y$. Then compare. $\endgroup$ – Rócherz Feb 26 '18 at 21:45
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The fact $x\geq1$ is a bit of a red Herring; I would try and do it under the weaker assumption $x\geq0$. Here is a start: $y>x$ implies we may write $y=x+\epsilon$ for some strictly positive number $\epsilon$. Substituting this in gives $$ y^2=(x+\epsilon)^2=\dots $$

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  • $\begingroup$ This is perfect, thank you! Will mark as correct when the system lets me. $\endgroup$ – Zaya Feb 26 '18 at 21:48
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Note that following inequalities are equivalent: $$ x^2 < y^2 $$ $$ x^2 - y^2 < 0 $$ $$ (x+y)(x-y) < 0 $$

And this is always true from our assumptions.

The first factor is positive: since $x \geq 1$ and $x<y$, then $x+y > 2 > 0$.
The second factor is negative: $x<y \Rightarrow x-y<0 $.
Product of two numbers, one of which is positive and the second negative, is always negative.

Now follow the inequalities bottom to up (they were equivalent) to arrive at the conclusion.

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Here is a simple way to go about it: You know that $y > x$, so $y - x > 0$. Now, since $x > 1$, both $x$ and $y$ are positive, $y + x > 0$. Can you finish it from there? (Doing it this way also tells you that even $x > 0$ is somewhat of a "weak" condition; all you need is $y + x > 0$, which should match your intuition about the square function: if $y > x$ and $y > -x$, then $y^2 > x^2$)

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I would suggest: $x = x \cdot 1 \leq \underbrace{x \cdot x}_{=x^2} < x \cdot y < y \cdot y = y^2$.

This is valid for all $x \in \mathbb{R}$, with $x < y$.

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For fun:

Let $x,y >0$, $x<y$, $x,y$ real.

Assume the contrary:

$x^2 \ge y^2$, then

$(x-y)(x+y)\ge 0.$

The second factor is positive, since x,y is positive.

Implies:

The first factor is non-negative, or

$x-y \ge 0,$ $x \ge y $,

a contradiction.

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