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anonymous

  • one year ago

Are any of the following statements is correct : i) 2 | a ⇒ 2| a^2 ii) 2 | a^2 ⇒ 2 | a iii) 2 | a ⇔ 2 | a^2 for integers a ?

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  1. anonymous
    • one year ago
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    They look all true.

  2. anonymous
    • one year ago
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    If you think a statement is false, try to produce a counterexample.

  3. anonymous
    • one year ago
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    @jayzdd but how do I show it?

  4. anonymous
    • one year ago
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    like in the first one\[a = 2*x \rightarrow a*a=2*x\] and because 2 is a divider to a it must be a divider to a*a right?

  5. nincompoop
    • one year ago
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    HA?

  6. anonymous
    • one year ago
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    @nincompoop that equal sign is if I should prime factorize a, and we know that 2 is a divider of a?

  7. anonymous
    • one year ago
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    @nincompoop could you explain instead?

  8. anonymous
    • one year ago
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    @welshfella

  9. amistre64
    • one year ago
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    (2n+1)(2n+1) = 4n^2+4n+1 = 2(2n^2+2n) + 1 = 2k+1 an odd squared, is odd but then since 2 doesnt divide odd numbers, it doesnt divide either factor to start with even*odd, divides the even number, and therefore the product and its square.

  10. anonymous
    • one year ago
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    @amistre64 could you explain a little more what your doing up there?

  11. amistre64
    • one year ago
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    just working some thoughts ... ------------------------ spose we have an even number, it is divisible by 2 if we multiply it by itself, we have 2 factors, each divisible by 2 ----------------------------- spose we have an odd number, it is NOT divisible by 2 if we multiply it by itself, we have 2 factors, NEITHER are divisible by 2 ------------------------------ spose we have a number that is the product of an even and an odd, it is divisible by 2 if we multiply it by itself, we have 2 factors that are still divisible by 2 ------------------------------- ergo if a number is divisible by 2, its square is divisible by 2 if a squared number is divisible by 2, then its primary value is also divisible by 2 this creates a bidirectional setup

  12. amistre64
    • one year ago
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    2n(2n) = 2(2n^2) 2n(2m+1) = 2(2nmk+n) = 2k, an even number odd number are not divisible by 2, so any combination of odd numbers will not be divisible by 2 to start with.

  13. amistre64
    • one year ago
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    i) 2 | a ⇒ 2| a^2 if a is even, then a^2 is even .... so true ii) 2 | a^2 ⇒ 2 | a since a^2 is even only if a is even ... this is also true iii) 2 | a ⇔ 2 | a^2 since i and ii are true, iii is correct by default.

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