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Thread: Proof - Implication signs

  1. #1
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    Proof - Implication signs

    Hi all,

    For these pairs of statements replace ... with $\displaystyle \Rightarrow \text{,} \Leftarrow \text{ or }\Leftrightarrow$. Assume $\displaystyle N$ is a positive Integer.

    The last digit of $\displaystyle N$ is 1 ... The last digit of $\displaystyle N^2$ is 1.

    thanks sammy
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  2. #2
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    (i) Consider $\displaystyle N=9$
    (ii) Assume the last digit of N is 1. That is, $\displaystyle N = 1(mod \ 10)$. Use the fact that $\displaystyle a \cdot b (mod \ n) = (a(mod \ n)\cdot b(mod \ n)) (mod \ n)$.
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  3. #3
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    Try to see if it works from $\displaystyle N$ to $\displaystyle N^2$ :

    $\displaystyle N \equiv 1$ (mod $\displaystyle 10$)

    So, $\displaystyle N = 1 + 10k$

    Now, $\displaystyle N^2 = (1 + 10k)^2$, which is equivalent to :

    $\displaystyle N^2 = 1^2 + 20k + (10k)^2$
    $\displaystyle N^2 = 1^2 + 20k + 100k^2$

    Simple deductions :
    $\displaystyle 20k \equiv 0$ (mod $\displaystyle 10$)
    $\displaystyle 100k^2 \equiv 0$ (mod $\displaystyle 10$)

    So, $\displaystyle N^2 \equiv 1$ (mod $\displaystyle 10$)

    Therefore, $\displaystyle N \Rightarrow N^2$ (not formally written but you understand)

    Now you try the other way round ($\displaystyle N^2$ to $\displaystyle N$), and see if it works. Then you will know what symbol to put ...
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  4. #4
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    That's great . This question is in the introduction to proofs in my 'AS' level book and we have not covered modulus yet. It was aimed really at using the correct implication sign.

    I could write out the first few terms and did see both ended in the digit one, but i couldnt work out how to give it a proof.

    I can understand why terms 11, 21, 31, 41 would result in $\displaystyle N=1(mod \ 10)$, and that the terms squared would also be $\displaystyle 1(mod \ 10)$ but does $\displaystyle 1 \neq 1(mod \ 10)$?

    sorry if this is dumb question.

    sammy
    Last edited by sammy28; Nov 12th 2009 at 08:49 AM. Reason: latex error
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  5. #5
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    Hi sammy

    $\displaystyle 1 \equiv 1 (mod \ 10)$
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