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Math Help - 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 \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 \sqrt{2} = \frac{x}{y} then the fraction cannot possibly be in reduced form.

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

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

    Thus x^2 is an even number.

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

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

    Thus y^2 is also an even number.

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

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

    Thus \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 e in Hardy
    and Wright.

    It gose like this:

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

    If k \ge b, and:

    ........... <br />
\alpha=k!\left(e-1-\frac{1}{1!}-\frac{1}{2!}- \cdots -\frac{1}{k!}\right)<br />

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

    ........... <br />
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} <br />

    (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 \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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