Question

Yaloswagomatics

Answer 1

@Shikanu

Answer 2

Theorem: there exists two irrational numbers n and m such that \(\large n^m\) is rational

Answer 3

let us use the number \(\Huge \sqrt{2}^\sqrt{2}\)

Answer 4

if \(\Huge \sqrt{2}^\sqrt{2}\) is irrational, then \(\Huge \sqrt{2}^{\sqrt{2}^\sqrt{2}}\)=\(\Huge \sqrt{2}^{\sqrt{2}*\sqrt{2}}=\Huge \sqrt{2}^2=2\)

Answer 5

\Huge \sqrt{2}^{\sqrt{2}^\sqrt{2}}
Where did this come from

Answer 6

well take n=\(\sqrt 2\) and take m=\(\Huge \sqrt{2}^\sqrt{2}\)

Answer 7

Okay

Answer 8

if \(\Huge \sqrt{2}^\sqrt{2}\) is rational, then take n=\(\sqrt{2}\) and m=\(\sqrt 2\) and you are finished.

Answer 9

QED

Answer 10

This is harder to wrap my head around than the other thereom.

Answer 11

so whether or not \(\Huge \sqrt{2}^\sqrt{2}\) is rational or irrational we just proved the theorem.

Answer 12

?
NO luck here

Answer 13

OK. Let's start over.
Theorem: there exists two irrational numbers n and m such that \(n^m\) is rational

Answer 14

Let us look at the number \(\Huge \sqrt{2}^\sqrt{2}\)
It is either rational or irrational, correct?

Answer 15

I thought it was just rational
I do not see how it is both

Answer 16

the number \(\Huge \sqrt{2}^\sqrt{2}\) is either rational or irrational (but not both) correct?

Answer 17

Oh
Ok

Answer 18

let us first assume that the number \(\Huge \sqrt{2}^\sqrt{2}\) is rational. Then, for the theorem we attempt to prove, take n=\(\sqrt 2\) and m=\(\sqrt 2\) and the theorem is true.

Answer 19

Correct?

Answer 20

Okay, so in the above example you said it was rational. What happens if you assume it is irrational. What changes?

Answer 21

@Gin_Ichimaru

Answer 22

Let us now assume that \(\Huge \sqrt{2}^\sqrt{2}\) is irrational. Then, \(\large (\sqrt{2})^{\sqrt{2}^\sqrt{2}}=(\sqrt{2})^{\sqrt2*\sqrt2}\)

Answer 23

=\(\sqrt2^2=2\)

Answer 24

take n=\(\sqrt{2}\) and m=\(\Huge \sqrt{2}^\sqrt{2}\) and the theorem is true.

Answer 25

therefore the theorem is true. QED

Answer 26

So since it is neither rational or irrational,
The answer will always be 2

Answer 27

?

Answer 28

No.

Answer 29

If \(\Huge \sqrt{2}^\sqrt{2}\) is rational then the theorem is true.
If \(\Huge \sqrt{2}^\sqrt{2}\) is irrational then the theorem is still true.

Therefore the theorem must be true.

Answer 30

I broke it down into two cases, and showed that for either case that the theorem was true.

Answer 31

this is called a non-constructive proof because I never established whether or not \(\Huge \sqrt{2}^\sqrt{2}\) is rational or irrational. I merely used it as an example, and showed that in any case the theorem would be true, if only I took different values for n and m.

Answer 32

Oh okay

Answer 33

Yoloswagomatics is a good name for this, no doubt

Answer 34

yup.