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    (Original post by TheJ0ker)
    Yeah I'm asking because I'm doing the STEP papers and the only formal proof I learn on Edexcel if proof by induction and they ask you about other methods of proving stuff on the STEP exams
    1. Have a look here.

    2. Have a read here.

    3. More to read here.


    These are good ones
    *. Yet more to read here.
    **. Yet more to read here 2.

    ... and these are all from Google ..
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    (Original post by gff)
    1. Have a look here.

    2. Have a read here.

    3. More to read here.


    These are good ones
    *. Yet more to read here.
    **. Yet more to read here 2.

    ... and these are all from Google ..
    Cheers

    Those are really really good!
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    (Original post by gff)
    \Rightarrow 2b - a = 2k for some k \in \mathbb{Z}

    ...


    \therefore hcf(a, b) \not= 1
    Could you please clarify what you did there? When I substitute 2b-2k for a, I end up with b=k.

    How does that lead to \therefore hcf(a, b) \not= 1?
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    (Original post by und)
    Could you please clarify what you did there? When I substitute 2b-2k for a, I end up with b=k.

    How does that lead to \therefore hcf(a, b) \not= 1?
    This is NOT a proof of anything, so don't even try to follow it. I should have made it clearer.
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    (Original post by gff)
    This is NOT a proof of anything, so don't even try to follow it. I should have made it clearer.
    Well, it's not a proof because there's no contradiction. However, Hasufel's proof has a clear contradiction: a and b are both even, therefore they have a common factor. What's wrong with that proof?
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    (Original post by und)
    Well, it's not a proof because there's no contradiction. However, Hasufel's proof has a clear contradiction: a and b are both even, therefore they have a common factor. What's wrong with that proof?
    Perhaps you did notice that I reduced my statement to the statement of Hasufel?

    What you are saying has nothing to do with my proposition.
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    (Original post by gff)
    Perhaps you did notice that I reduced my statement to the statement of Hasufel?

    What you are saying has nothing to do with my proposition.
    And what exactly is your proposition? I'm now quite confused as you can probably tell.

    Hence, by your argument 2 - \sqrt{4} = 2 - 2 = 0 is an irrational number.
    Should I ignore this too? Whose argument are you referring to?
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    (Original post by und)
    And what exactly is your proposition? I'm now quite confused as you can probably tell.



    Should I ignore this too? Whose argument are you referring to?
    I don't see why we should waste more time with this?

    2 - \sqrt{16} = \frac{a}{b}, hcf(a, b) = 1

    \Rightarrow 2b - a = b\sqrt{16}

    \Rightarrow (2b - a)^2 = 16b^2

    \Rightarrow 2b - a = 2k, k \in \mathbb{Z}

    .. from now on you can continue with the proof given above.


    What I'm asking is, what makes it different? Probably that you know in advance that \sqrt{2} is irrational? (what you pretend you aren't assuming)

    Reduce it to contradiction and let me know.
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    (Original post by gff)
    I don't see why we should waste more time with this?

    2 - \sqrt{16} = \frac{a}{b}, hcf(a, b) = 1

    \Rightarrow 2b - a = b\sqrt{16}

    \Rightarrow (2b - a)^2 = 16b^2

    \Rightarrow 2b - a = 2k, k \in \mathbb{Z}

    .. from now on you can continue with the proof given above.


    What I'm asking is, what makes it different? Probably that you know in advance that \sqrt{2} is irrational? (what you pretend you aren't assuming)

    Reduce it to contradiction and let me know.
    It's quite different, isn't it? If you continue with your 'proof', there's no contradiction, whereas when you're trying to prove that 2-\sqrt2 is rational you encounter a clear contradiction. Therefore, I don't see what you're getting at. In fact, one should assume that that 2-\sqrt2 is rational until finding the contradiction.

    I'm sorry if I've misunderstood you.
 
 
 
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