True or False: $2^{2^{2011}} \text{ divides } 2^{2^{2012} }$

Question

True or false: $$2^{2^{2011}} \text{ divides } 2^{2^{2012} }$$

Give your justifications. I don't know how to start this problem so far.

But, I guessed like this, $$2^{\underbrace{2\times 2\times\ldots\times 2}_{2011 times}}\text{ divides } 2^{\underbrace{2\times 2\times\ldots\times 2}_{2012 times}}$$

$$\implies \qquad \underbrace{2\times2\times\ldots\times 2}_{2^{2011}\;times}\text{ divides } \underbrace{2\times2\times\ldots\times 2}_{2^{2012}\;times}$$

Number of factors in the first product is less than that of second product. So defenitly,it will divide.

Kindly say is my reason is sound? If its ok, then give me alternate details about if we change the base . $\left(\text{That is what about }a^{2^{2011}}\right.$ and $\left.b^{2^{2012}}\right)$ where $gcd(a,b)=d\neq 1$

Answer 1

The reasoning here is very simple. $2^n$ divides $2^m$ for $n,m$ positive integers if and only if $n \le m$ ; this is just because you multiply a bunch of $2$'s together, so if there are less multiples of $2$ on one side than on the other, then the side with less $2$'s divides the other side.

Now $n=2^{2011} \le 2^{2012}=m$ shouldn't be hard to see.

Hope that helps,

Answer 2

More specifically, $2^{2^{2012}}=2^{2^{2011}\times2}=(2^{2^{2011}})^{2}$

Answer 3

True because $$2011 +1 = 2012 \Longrightarrow 2^{2012} = 2 \cdot 2^{2011} \Longrightarrow 2^{2^{2012}}= (2^{2^{2011}})^2$$