Question

If a set has a maximum value, then there exists a number in that set such that it's greater or equal to all elements of that set.

Answer 1

Prove
\[A = \left\{ p \in Q:p>0 \cap p^{2}<2 \right\}\] has no maximum

Answer 2

Then
\[\sqrt{2} = p + (\sqrt{2}-p) = p+\frac{(\sqrt{2}-p)(\sqrt{2}+p)}{(\sqrt{2}+p)}\]
\[=p+\frac{2 - p^{2}}{\sqrt{2}+p} > p+\frac{2-p^{2}}{2 + p}\]

Answer 3

Let

\[q = p+\frac{2-p^{2}}{2 + p}\]
Then we have proved q > p.
Now we need to prove q^2 < 2.

\[q^{2} = (p+\frac{2-p^{2}}{2+p})^{2} = (\frac{2p + p^{2}+2-p^{2}}{2+p})^{2} = (\frac{2(p+1)}{2+p})^{2}\]
\[=\frac{4(p+1)^{2}}{(2+P)^{2}}\]
We must prove this is less than 2
\[\frac{4(p+1)^{2}}{(2+P)^{2}} < 2\]
\[\frac{2(p+1)^{2}}{(2+p)^{2}} < 1\]
\[2(p+1)^{2} < (2+p)^{2}\]
\[2p^{2}+4p+2 < p^{2} + 4p + 4\]
\[p^{2} < 2\]
Thus we have proved that q^2 is less than 2

Answer 4

I can't check your ideology as I haven't taken linear algebra but your algebraic computation looks right.

Answer 5

This isn't linear algebra xd, it's just math..

Answer 6

lmao I didn't even pay attention, I suppose I was under the impression it would be a question involving topics I'm not familiar with.

Answer 7

Your math does look correct. It seems you have followed everything correctly.

Answer 8

Imagine trying to memorize this only to get a different question on the test xd

Answer 9

Hey, I get that all the time. I've spent like so much time remembering equations and such, only to get to the test and need like the two or three I didn't.