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Math Help - Proof - Implication signs

  1. #1
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    Proof - Implication signs

    Hi all,

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

    The last digit of N is 1 ... The last digit of N^2 is 1.

    thanks sammy
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  2. #2
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    (i) Consider N=9
    (ii) Assume the last digit of N is 1. That is, N = 1(mod \ 10). Use the fact that 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 N to N^2 :

    N \equiv 1 (mod 10)

    So, N = 1 + 10k

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

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

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

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

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

    Now you try the other way round ( N^2 to 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 N=1(mod \ 10), and that the terms squared would also be 1(mod \ 10) but does 1 \neq 1(mod \ 10)?

    sorry if this is dumb question.

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

    1 \equiv 1 (mod \ 10)
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