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I'm working on a basic proof for my intro proof course. The text is Analysis with an Introduction to Proof by Lay. This question comes from section 2. I am asked to proof or disprove the following:

If $x>5$, then there exists a $y>0$ such that $x^2>25 + y$

So my work follows:

Proof: Suppose $x$ is an arbitrary real number greater than 5 and choose $y=x-5$, then

$25 + y = 25 + (x-5)$

$=20+x$

Which is less than $x^2$ when $x>5$. Therefor for all $x>5$ there exists a $y$ namely, $y=x-5$ such that $x^2>y+25$

So that's what I have. At this point I'm not entirely convinced that I've shown enough to proove my statement. I'm still very green when it comes to writing proofs and I feel as if my conculsion is not obvious. I think I'm on the right track here but I could use some direction. Also I've been scolded more than once for doing proofs backwards, I'm hoping that I've alleviated this in this problem but again, this is all still pretty new. Any help is appreciated.

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    $\begingroup$ Take rather $z=x-5>0$. Then expand $x^2=(5+z)^2$. You'll see that you can take $y=z^2>0$. $\endgroup$ – Julien Feb 8 '13 at 15:22
  • $\begingroup$ Your proof is correct. But note that finding any one value of y which satisfies the condition is sufficient for existance proofs. $\endgroup$ – user45099 Feb 8 '13 at 16:05
  • $\begingroup$ Right, the existential quantifer vs. the universal quantifier correct? I only need to find 1 y which satisfies for all x... $\endgroup$ – Ben Anderson Feb 8 '13 at 16:33
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Your proof is correct, but you should elaborate on why $x^2>20+x$. This inequality is the same as $(x-5)(x+4)=x^2-x-20>0$ which is true when $x>5$.

Alternatively, just use $y=\frac{x^2-25}{2}$ to lead to a quick proof.

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  • $\begingroup$ can you possibly elaborate on this, I think I see why this works, however, my professor was not satified with the above - that is, I had not sufficiently shown why (in my proof) that 20+x is less than $x^2$. I can see why taking the above works, I just can't articulate it correctly I guess. $\endgroup$ – Ben Anderson Feb 8 '13 at 16:35

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