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

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

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

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

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

    Step 4:
    $\displaystyle 2b^2 = (2k)^2$
    $\displaystyle 2b^2 = 4k^2$
    $\displaystyle b^2 = 2k^2$
    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 $\displaystyle \sqrt{2}$ 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 $\displaystyle \sqrt{3}$ example.


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

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

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

    Step 3: $\displaystyle 3b^2 = a^2$
    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:
    $\displaystyle 3b^2 = (2k + 1)^2$
    ......

    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 $\displaystyle \displaystyle 3b^2 = a^2$. So $\displaystyle \displaystyle a^2$ is a multiple of $\displaystyle \displaystyle 3$. This means $\displaystyle \displaystyle a$ is a multiple of $\displaystyle \displaystyle 3$. This can be proven using the contrapositive.

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

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

    $\displaystyle \displaystyle 3b^2 = 9k^2$

    $\displaystyle \displaystyle b^2 = 3k^2$.

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

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

    this can be used to prove $\displaystyle \sqrt{3}$ 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 $\displaystyle a^2$ a multiple of 3 implies $\displaystyle a$ is a multiple of 3 (by contrapositive).

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