## 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 ?

1. anonymous

They look all true.

2. anonymous

If you think a statement is false, try to produce a counterexample.

3. anonymous

@jayzdd but how do I show it?

4. anonymous

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

HA?

6. anonymous

@nincompoop that equal sign is if I should prime factorize a, and we know that 2 is a divider of a?

7. anonymous

8. anonymous

@welshfella

9. amistre64

(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

@amistre64 could you explain a little more what your doing up there?

11. amistre64

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

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

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.