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Math Help - Rational Number

  1. #1
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    Rational Number

    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?
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  2. #2
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    Quote Originally Posted by JimmyT View Post

    My question is, what theorem, reasoning, or whatever allows for this to be true and do you think this is adequate for a proof?
    It uses the rational root test.
    The reason of the proof is by contradiction.
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  3. #3
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    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

    . . . . . . . . . . . . . . . . . . _ . . . .
    Square both sides: . 2 + √2 . = . ----
    . . . . . . . . . . . . . . . . . . . . . . . .

    . . . . . . . . . . . ._ . . . .
    And we have: . √2 . = . --- - 2 . [1]
    . . . . . . . . . . . . . . . . .

    . . . . a . . . . . . . . . . . .
    Since -- is rational, then --- is rational. .(Integers are closed under mutliplication)
    . . . . b . . . . . . . . . . . .

    . . . . . . . .
    . . Hence, --- - 2 is rational. .(Rational numbers are closed under subtraction.)
    . . . . . . . .


    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.

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  4. #4
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    Quote Originally Posted by Soroban
    Integers are closed under mutliplication
    You mean rationals.
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  5. #5
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    Hello, TPHacker!

    You mean rationals.
    Well, I was thinking that aČ is an integer and bČ is an integer,
    . . which makes aČ/bČ a rational number.

    But you're right ... The closure of rationals is a much better reason.
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