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Math Help - Prove that a+b is irrational?

  1. #1
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    Prove that a+b is irrational?

    okay so the question is: Suppose that a is a rational number and that b is an irrational number. Prove that a+b is irrational.

    So i went to prove A: ab is irrational by contradiction, assuming A is false.

    Therefore ab is rational, so ab = m/n for integers m and , n is nonzero. So if a is given to be rational then a = k/l for integers k and l, l and k are both nonzero.

    So b = m/n + a .... then i completely got lost...

    Could someone help me from there, or let me know if I am completely wrong? Thanks.
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  2. #2
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    Quote Originally Posted by GreenDay14 View Post
    okay so the question is: Suppose that a is a rational number and that b is an irrational number. Prove that a+b is irrational.

    So i went to prove A: ab is irrational by contradiction, assuming A is false.

    Therefore ab is rational, so ab = m/n for integers m and , n is nonzero. So if a is given to be rational then a = k/l for integers k and l, l and k are both nonzero.

    So b = m/n + a .... then i completely got lost...

    Could someone help me from there, or let me know if I am completely wrong? Thanks.
     a is rational and  b is irrational.

    Therefore  a = \frac{p}{q} .

    Now, let's assume that  a+b = \frac{l}{m} , i.e., assume that a+b is rational.

    Now, since  a = \frac{p}{q} , we can write  a+b = \frac{p}{q} + b

    Therefore  \frac{p}{q} + b = \frac{l}{m}

    Therefore  b = \frac{l}{m} - \frac{p}{q} = \frac{ql - pm}{mq}

    This is a contradiction since  \frac{ql - pm}{mq} is rational, and b was defined as irrational. Therefore, the assumption that a+b is rational must be false, and a+b must be irrational.
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