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If $2$ divides a number $a$, does $2^n$ divide $a$ ? $n$ is any integer.

This seems to be true for me, but I just want to make sure it applies for all numbers.

example

if a = 137

2 does not divide 137 4 does not divide 137 8 does not divide 137 and so on

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    $\begingroup$ What multiple of what divides what? The way I read this you claim that if $2 \mid a$, then $2n \mid a$ for every integer $n$. That's clearly wrong. $\endgroup$
    – fgp
    Apr 11, 2014 at 10:44
  • $\begingroup$ What happens if $a=2$? Or are you wanting to ask about $2$ dividing multiples of $a$? $\endgroup$ Apr 11, 2014 at 10:44
  • $\begingroup$ hi thank you for the respond I have changed the question please read it now. $\endgroup$
    – user142405
    Apr 11, 2014 at 11:01
  • $\begingroup$ No. 2 divides 6, but $8=2^3$ does not divide 6. $\endgroup$
    – ah11950
    Apr 11, 2014 at 11:07
  • $\begingroup$ $2$ divides $2$ but $2^2$ does not :) $\endgroup$
    – mesel
    Apr 11, 2014 at 11:13

3 Answers 3

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$2$ divides $4$

$2\cdot 4=8$ does not divide $4$

Over and Done!!

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Edit: question was changed again, my answer is no longer valid.

Yes, if a divides b then a divides the multiples of b.

Consider b/a = c, where c is a whole number. Take any number d, and multiply it by b. This gives us bd. By our formula early, this means bd/a = dc, where dc is the product of two whole numbers 9i.e, another whole number). This is true for any a, b and d where a divides b.

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Do you now mean

does $2^n$ divide $a^n$?

Let $a=2b$, then $a^n=(2b)^n=2^nb^n$

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  • $\begingroup$ what I mean is basically this let a= 137 2 does not divide 137 2 x 2 = 4 does not divide 137 2x2x2 = 8 does not divide 137 seems like 2^n doesn't divide 137 $\endgroup$
    – user142405
    Apr 11, 2014 at 11:18
  • $\begingroup$ @user142405 That is a completely different question. You have observed that if $a$ is odd (another way of saying that $2$ does not divide $a$) and $n\ge 1$ then $2^n$ does not divide $a$ - suppose it did so that $a=2^nb$ then $a=2\cdot 2^{n-1}b$. $\endgroup$ Apr 11, 2014 at 13:08

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