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Math Help - to show there are irrational numbers

  1. #1
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    to show there are irrational numbers

    hi everyone.
    if we have to prove there are irrational number x and y are the element of Real numbers,
    for example 2x-y is rational (by contradiction).
    so how would that be? just getting two random numbers and substitute them into the equation? cant work out what it is so please, can anyone help me? thankyou so much!
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  2. #2
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    Hello yanyannn

    Welcome to Math Help Forum!
    Quote Originally Posted by yanyannn View Post
    hi everyone.
    if we have to prove there are irrational number x and y are the element of Real numbers,
    for example 2x-y is rational (by contradiction).
    so how would that be? just getting two random numbers and substitute them into the equation? cant work out what it is so please, can anyone help me? thankyou so much!
    I'm sorry, but you must write in intelligible English if we are to understand what you mean. This just doesn't make sense.

    Please set out what you mean as carefully and as precisely as you can.

    Grandad
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  3. #3
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    Well

    Granddad is much better in both explaining and understanding/answering math questions then me.
    So explaining to him what you mean is a good idéa.
    However:
    I understand your question to be:
    Proving that there are irrational numbers by contradiction. Then I assume that the example you made (2x-y) is your own.

    I might be wrong but, substituting random number as x;y doesnt give you anything. You have to substitute an irrational number and then prove that it can not be written as rational. (that is in the form of \frac{a}{b})
    A classical example is \sqrt{2} There is a irrational number, and the proof very old, checking that proof could give you an idéa, (it is done by contradiction).
    Hope it was helpfull...
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