I have the following problem:
Prove that the sqrt(2+sqrt(2)) is irrational.
I did the problem similarly to how one of my TAs showed in a tutorial.
Let the sqrt(2+sqrt(2)) = a. Square both sides. This is:
2+sqrt(2)=a
sqrt(2)=(a^2)-2 Square both sides
2=(a^4)-(4a^2)+4
0=(a^4)-(4a^2)+2
Because the factors of 2 (namely 1, -1, 2, -2) do not solve the following equation, we can assume that the original function is irrational.
My question is, what theorem, reasoning, or whatever allows for this to be true and do you think this is adequate for a proof?
Hello, JimmyT!
I would try to prove it by contradiction . . .
Prove that the sqrt(2+sqrt(2)) is irrational.
Assume that expression is rational.
. . . . . . .______. . . a
That is, √2 + √2 .= .--- . for some integers a and b.
. . . . . . . . . . . . . . .b
. . . . . . . . . . . . . . . . . . _ . . . . aČ
Square both sides: . 2 + √2 . = . ----
. . . . . . . . . . . . . . . . . . . . . . . .bČ
. . . . . . . . . . . ._ . . . . aČ
And we have: . √2 . = . --- - 2 . [1]
. . . . . . . . . . . . . . . . .bČ
. . . . a . . . . . . . . . . . . aČ
Since -- is rational, then --- is rational. .(Integers are closed under mutliplication)
. . . . b . . . . . . . . . . . . bČ
. . . . . . . .aČ
. . Hence, --- - 2 is rational. .(Rational numbers are closed under subtraction.)
. . . . . . . .bČ
Hence, the right side of equation [1] is a rational number.
. . But the left side is irrational.
We have reached a contradition; our assumption is incorrect.
. . . . . . . . . ______
Therefore: . √2 + √2 .is irrational.