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Thread: How to prove square root 2 is irrational?

  1. #1
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    How to prove square root 2 is irrational?

    I have the first bit to this problem but I do not know how to prove that x^2 is even and then y is even. I am using the contradiction method. Here is what I have so far :

    Assume sqrt2 is rational

    therefore sqrt2=(x/y)^2
    2y^2=x^2

    thanks for any help.

    p.s how do you write with apprioate symbols? I have never done maths on the computer.
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  2. #2
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    Quote Originally Posted by Benno_11 View Post
    I have the first bit to this problem but I do not know how to prove that x^2 is even and then y is even. I am using the contradiction method. Here is what I have so far :

    Assume sqrt2 is rational

    therefore sqrt2=(x/y)^2
    2y^2=x^2

    thanks for any help.

    p.s how do you write with apprioate symbols? I have never done maths on the computer.
    We have to state something slightly stronger than $\displaystyle \frac{x}{y}$ is rational, we have to state that it has been reduced to lowest terms. We may freely make this assumption. The point of the proof is to show that if $\displaystyle \sqrt{2} = \frac{x}{y}$ then the fraction cannot possibly be in reduced form.

    So assume we can write
    $\displaystyle \sqrt{2} = \frac{x}{y}$
    where x and y are integers ($\displaystyle y \neq 0$) and the fraction is in reduced form. Then by squaring both sides we see that:
    $\displaystyle \frac{x^2}{y^2} = 2$

    $\displaystyle x^2 = 2y^2 = 2(\text{integer})$

    Thus $\displaystyle x^2$ is an even number.

    So if x is an even number it is of the form $\displaystyle x = 2n$ where n is an integer. Thus $\displaystyle x^2 = 4n^2$ and so
    $\displaystyle 2y^2 = 4n^2$

    $\displaystyle y^2 = 2n^2 = 2(\text{integer})$

    Thus y^2 is also an even number.

    Thus y is of the form $\displaystyle y = 2m$ where m is a (non-zero) integer.

    Thus
    $\displaystyle \frac{x}{y} = \frac{2n}{2m}$
    which can be reduced, contrary to assumption.

    Thus $\displaystyle \sqrt{2}$ is not rational.

    -Dan
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    thanks

    Thanks for your help

    You could also use this for sqrt 3, couldn't you?

    Matt

    P.S
    I like your user name and your quote
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    Quote Originally Posted by Benno_11 View Post
    You could also use this for sqrt 3, couldn't you?
    Yes
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    Quote Originally Posted by Benno_11 View Post
    Thanks for your help

    You could also use this for sqrt 3, couldn't you?

    Matt

    P.S
    I like your user name and your quote
    It works for the square root of every positive integer that is not a
    perfect square.

    RonL
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    How would you ...

    How do you prove that e and pi are irrational? I do not need to know it, but I was just wondering how complex it would be. For e, would you have to use the maclaurin series?
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    Quote Originally Posted by Benno_11 View Post
    How do you prove that e and pi are irrational? I do not need to know it, but I was just wondering how complex it would be. For e, would you have to use the maclaurin series?
    There is a neat, short proof of the irrationality of $\displaystyle e$ in Hardy
    and Wright.

    It gose like this:

    Let's suppose that $\displaystyle e$ is rational, so that $\displaystyle e=a/b$ and $\displaystyle a$ and $\displaystyle b$ are integers.

    If $\displaystyle k \ge b$, and:

    ........... $\displaystyle
    \alpha=k!\left(e-1-\frac{1}{1!}-\frac{1}{2!}- \cdots -\frac{1}{k!}\right)
    $

    then $\displaystyle b|k!$ and $\displaystyle \alpha$ is an integer. But (replacing $\displaystyle e$ by its series expansion)

    ........... $\displaystyle
    0<\alpha=\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}+ \cdots < \frac{1}{k+1}+\frac{1}{(k+1)^2}+\cdots = \frac{1}{k}
    $

    (the last part of the above uses the sum of the geomentric series)

    RonL
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    Quote Originally Posted by Benno_11 View Post
    How do you prove that e and pi are irrational? I do not need to know it, but I was just wondering how complex it would be. For e, would you have to use the maclaurin series?
    The proof that $\displaystyle \pi$ is irrational is much more complicated. The first proof was published by the mathematician Lambert.

    Here.

    Here is a good link devoted to Real Analysis. It does a better job explanaing the proof.
    Here.
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  9. #9
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    squareroot-of-2

    Quote Originally Posted by Benno_11 View Post
    I have the first bit to this problem but I do not know how to prove that x^2 is even and then y is even. I am using the contradiction method. Here is what I have so far :

    Assume sqrt2 is rational

    therefore sqrt2=(x/y)^2
    2y^2=x^2

    thanks for any help.

    p.s how do you write with apprioate symbols? I have never done maths on the computer.
    Here both x and y are relatively prime nos(i.e they don"t have a common factor.)then their squares are also relatively prime. Now we have 2y^2=x^2
    therefore y^2=x^2/2 which is a contradiction that x^2 and y^2 are realtivelyprime.I hope this may clear your doubt.
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