Why does my calculator show $2^{-329} = 0?$

Question

On my calculator, I usually get a $0$ when I divide something by $2$, a lot of times if that makes sense, but I was just wondering why does $2^{-329} = 0?$

Answer 1

The precise name for the feature is underflow, which means it's less than the smallest number the registers can hold. It's kind of like overflow.

Answer 2

It doesn't. Your calculator can't handle a number of such a small magnitude. Specifically, $2^{-329}\approx 9.14\times 10^{-100}$, and I'm guessing that your calculator can only handle numbers of magnitude between $10^{-99}$ and $10^{99}$.

Answer 3

Your calculator gives you $0$ for $2^{-329}$ for the same reason that it gives you $0$ when you divide by $2$ a lot: Because $2^{-329}$ is dividing by $2$ a lot. More precisely,

$$2^{-329}=\dfrac{1}{2^{329}}=\dfrac{1}{\text{huge number}}=\text{number so small your calculator doesn't know it from }0.$$

Note that $2^{-329}$ is what you get if you start with $1$ and divide by $2$ repeatedly, a total of $329$ times.

Answer 4

The calculator doesn't have enough precision to store the number which is so small that the closest number it can represent is 0.

Similarly, the same thing happens when you try repeatedly square rooting a number on your calculator.

I.e. pick any positive number $n$ and evaluate $\sqrt{\sqrt{\sqrt{...\sqrt{\sqrt{n}}}}}$. You'll find the calculator eventually returns $1.0$ when you do it enough times.

Answer 5

You know from your algebra course (hopefully) that $2^x \neq 0$ for any $x$, so this has something to do with calculator doing something in appropriate. You were probably expecting something in scientific notation: $$ 2^{-329} = a \cdot 10^b $$ for some $1 \leq a < 10$ and an integer $b$ (so that neither the right hand side nor the left hand side are zeroes, of course). Can you still compute this on your calculator? In your algebra class, you should have learned a trick to make insanely big (or insanely small, like this one) numbers manageable. Let's put this trick to action: $$ b + \log_{10} a = \log_{10} 2^{-329} = -329 \log_{10} 2 = -329 \cdot 0.30103 = -99.038869 $$ If $1 \leq a < 10$, then $0 \leq \log_{10} a < 1$. So $b$ is the integer floor part of the answer, and $\log_{10} a$ is the fractional part: $$ b = -100; \log_{10} a = 1 - 0.038869 = 0.961131; a = 10^{0.0.961131} = 9.14389 $$ Thus, $$ 2^{-329} = 9.14389 \cdot 10^{-100} $$ as of course pointed out by everybody else in this thread.

Calculator is your (powerful) tool, but you still need to be smart in using that tool.

Answer 6

Because, the calculator has a limit, and the input of yours crosses that limit. Your calculator calculates the value for the positive index first and then inverses it. Hence, it's underflowing its register values.