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Math Help - proving √q is irrational by contradiction

  1. #1
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    Exclamation proving √q is irrational by contradiction

    Let q≥2 be a positive integer. If for all integers a and b, whenever q|ab, q|a or q|b, then √q is irrational.

    I know I should do a proof by contradiction so i should assume √q is rational, and that for all integers a and b if q|ab then q|a or q|b.

    So since √q is rational it can be written as m/n where m and n are integer and are relatively prime.

    So √q=m/n. Squaring both sides q=m^2/n^2. Then q(n^2)=m^2.

    I got stuck at this point. I think im suppose to show q|m and q|n to show a contradiction but im not sure how to get there.

    Anything helps!! thanks!
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  2. #2
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    Quote Originally Posted by junebug5389 View Post
    So √q=m/n. Squaring both sides q=m^2/n^2. Then q(n^2)=m^2.

    I got stuck at this point. I think im suppose to show q|m and q|n to show a contradiction but im not sure how to get there.
    That's right, just do it: your equation shows that q divides m^2=m\times m, hence q divides m. Write m=kq to conclude that it divides n as well.
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  3. #3
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    maybe im missing a simple definition but i was not sure how to show since q|m^2 that q|m. or is it just implied that since q|m^2 that q|m and thats all i have to say. sorry this might be a stupid question...
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  4. #4
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    Quote Originally Posted by junebug5389 View Post
    maybe im missing a simple definition but i was not sure how to show since q|m^2 that q|m.
    This is because of your assumption on q (with a=b=m, that's why I wrote m^2=m\times m).

    By the way, this assumption is equivalent to saying that q is prime.
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  5. #5
    ddt
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    Quote Originally Posted by junebug5389 View Post
    Let q≥2 be a positive integer. If for all integers a and b, whenever q|ab, q|a or q|b, then √q is irrational.

    I know I should do a proof by contradiction so i should assume √q is rational, and that for all integers a and b if q|ab then q|a or q|b.
    Proof by contradiction is only needed partly.

    If the premise holds, i.e., if for all a and b: q | ab => q|a or q|b, then q is a prime. For suppose that q is not prime; then you can take a as one of the prime factors of q, and b = q/a.

    From that, it's trivial to see √q is irrational - the proof for any prime q is the same as the classical proof that √2 is irrational.
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