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Math Help - Proving a number is irrational by contradiction

  1. #1
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    Proving a number is irrational by contradiction

    Hi

    I have this worked example to prove \sqrt{2} is an irrational number via contradiction. I will put it down here now. But my question is how do I do this with \sqrt{3} as I get stuck half way through this when we come to talking about even numbers.

    Step 1: Suppose that \sqrt{2} is rational.

    Step 2: This means that its possible to find  a, b \in \mathbb{Z} where a and b have only a common factor of 1 such that:

    \sqrt{2} = \frac{a}{b}, b not equal to 0.
    Squaring both sides: 2 =\frac{a^2}{b^2}

    Step 3: 2b^2 = a^2
    This means that a^2 must be even and thus a must be even.
    i.e. a = 2k

    Step 4:
    2b^2 = (2k)^2
    2b^2 = 4k^2
    b^2 = 2k^2
    This means that b^2 and b must be even.

    Step 5: So both a and b are even. This means that a and b have a common factor or 2. This contradicts the original hypothesis and so the hypothesis must be false. This means that \sqrt{2} is not able to be expressed in the form a/b and is therefore an irrational number.

    The end.
    So that is the example proof. Here is what I have done so far with the \sqrt{3} example.


    Step 1: Suppose that \sqrt{3} is rational.

    Step 2: This means that its possible to find  a, b \in \mathbb{Z} where a and b have only a common factor of 1 such that:

    \sqrt{3} = \frac{a}{b}, b not equal to 0.
    Squaring both sides: 3 =\frac{a^2}{b^2}

    Step 3: 3b^2 = a^2
    This condition is only met when a is odd and b is odd. I just did this with a little trial and error. So a is odd. Thus a = 2k + 1

    Step 4:
    3b^2 = (2k + 1)^2
    ......

    Ok this is where I come unstuck. Does anyone have any suggestions?

    David.
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  2. #2
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    From Step 3, you have \displaystyle 3b^2 = a^2. So \displaystyle a^2 is a multiple of \displaystyle 3. This means \displaystyle a is a multiple of \displaystyle 3. This can be proven using the contrapositive.

    So we can let \displaystyle a = 3k and therefore

    \displaystyle 3b^2 = (3k)^2

    \displaystyle 3b^2 = 9k^2

    \displaystyle b^2 = 3k^2.

    Since \displaystyle b^2 is a multiple of \displaystyle 3 , so is \displaystyle b.

    There is your common factor and contradiction.
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  3. #3
    Senior Member BAdhi's Avatar
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    Quote Originally Posted by Bucephalus View Post
    This means that a^2 must be even and thus a must be even.
    i.e. a = 2k
    we can come to the same conclusion by p|a^2\implies p|a where p is a prime

    this can be used to prove \sqrt{3} is not a rational number since 3 is also a prime
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  4. #4
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    Thankyou both for your help

    Cheers.
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  5. #5
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    Here is a proof that a^2 a multiple of 3 implies a is a multiple of 3 (by contrapositive).

    Suppose a is not a multiple of 3. Then a=3k+1 or a=3k=2 for some integer k. If a=3k+1, then a^2=(3k+1)^2=9k^2+6k+1=3(3k^2+2k)+1. 3k^2+2k is an integer because the integers are closed under addition and multiplication. So a^2 is not divisible by 3. The second case is similar.
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