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Math Help - Very easy proof

  1. #1
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    Very easy proof

    I have done this one before can't remember how I went about it.

    Let x be a real number. Show that if x^2 is irrational, then so is x.

    So by contradiction:
    x^{2} is irrational and x is rational.

    A rational number squared won't be irrational but how do I show it?

    Thanks.
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  2. #2
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    Quote Originally Posted by dwsmith View Post
    Show that if x^2 is irrational, then so is x.
    Can you show that \text{if }x\text{ is rational then }x^2\text{ is rational}~?.
    If so, what is the contrapositive ?
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  3. #3
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    Quote Originally Posted by Plato View Post
    Can you show that \text{if }x\text{ is rational then }x^2\text{ is rational}~?.
    If so, what is the contrapositive ?
    If x is rational, then x=\frac{m}{n}, \ n\neq 0, \ m,n\in\mathbb{Z}.

    x^2=\frac{m}{n}\cdot\frac{m}{n}

    And a rational times a rational number is rational. The contrapositive is my original statement.

    Is that just it?
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  4. #4
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    The fact that the product of 2 rationals is rational follows from the fact that the product of two integers is an integer, and the product of two nonzero integers is nonzero.

    And your proof is really by contrapositive, not contradiction.
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  5. #5
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    Quote Originally Posted by dwsmith View Post
    If x is rational, then x=\frac{m}{n}, \ n\neq 0, \ m,n\in\mathbb{Z}.
    x^2=\frac{m}{n}\cdot\frac{m}{n}
    And a rational times a rational number is rational.
    The contrapositive is my original statement.
    Quote Originally Posted by DrSteve View Post
    And your proof is really by contrapositive, not contradiction.
    I think that he said as much.
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