## hana88 3 years ago if x is an integer divisible by 3, then x^2 is divisible by 3. is this true or false? is the converse true or false?

1. hana88

can any one help?

2. just.chris

x = N x 3, represents the divisible by 3 integer. x^2 then equals N^2 x 3^2 which having the factors of 3 is therefore divisible by 3. Is this useful?

3. hana88

yes thank you

4. hana88

what about the converse?

5. lopus

X=2*3 6/3=2 \[x^{2}=36\] 36/3=12

6. hana88

that only gives me evidence

7. just.chris

I haven't done this stuff in awhile. Do you mean if and integer, x, is divisible by 2 then x^3 is also divisible by 2?

8. hana88

no if x^2 is divisive by 3, then x is divisible by 3

9. just.chris

OK. I am winging it here In order for x^2 to be divisible by a prime, which cannot be further factored then it must also be a factor of x. Is this useful?

10. lopus

x=n*3 n=1,2,3,4,5,6,7,8............... x^2=3,36,81,144....... divide

11. hana88

@lopus x is 3,6,9,12,18...

12. just.chris

x = 3,6,9,... is correct valus for x but does that indicate that if x^2 is divisible by 3 then so is x? I am asking because I don't know. A question about the question.

13. lopus

yes, you just divide and see for yourself demostration x=N*3 (N*3)^2 =n^2 * 3^3

14. lopus

excuse me 3^2

15. just.chris

So you don't really need the series you just need if x^2 is divisible by 3 then x is because 3 cannot be further factored. And the equations you provided apply in bothe directions. unlike say if x^2 is divisible by 4 then x is not necessarily divisible by for because 4 can be further factored, i.e., 2^2 = 4 is divisible by 4 but x, being 2, is not. Yes?

16. Mertsj

If x^2 is divisible by 3 it can be written in the form 3a where a is an integer. So x^2=3a and x = sqrt3(sqrt a) so the converse is not true.

17. helder_edwin

3 is a prime number. u should pay attention to that!!!

18. hana88

um @helder you make no sense

19. Mertsj

Why?

20. hana88

@Mertsj thanks thats the best answer because it was easy to understand

21. helder_edwin

if \(3\mid x^2\) then \(3\mid x\) because 3 is prime.

22. Mertsj

yw

23. Mertsj

That is not true helder. 3|15 but 3 does not divide sqrt15

24. Mertsj

So if x^2 = 15, 3 is a factor of x^2 but not a factor of x

25. helder_edwin

but 15 is not a perfect square!

26. Mertsj

The converse is: If x^2 is divisible by 3 then x is an integer divisible by 3. So we see that x^2 does not have to be a perfect square. It only has to be divisible by 3.

27. helder_edwin

so why it is x^2 if it is not a perfect square?

28. Mertsj

x^2 could be 15 or 18 or 21 or 24 or 27 or 12 .

29. helder_edwin

nerver mind.

30. just.chris

If x^2 does not have to be a perfect square then the converse is not true. However, if x^2 is perfect and divisible by 3 then is the converse true? And would that be the case for any prime? Since primes cannot be factored into smaller integers then I think in order for the square to be factorable by the prime then you must have the prime^2 in order to have the prime as a factor of x^2.