In trying to relearn scientific notation after years since I left school, I noticed that when we get a very small number and convert it to scientific notation, you're actually multiplying the tiny number a certain number of times so that the number becomes something between 1 and 10, and you write the inverse equation as scientific notation. Example: 0.0035

I multiply it by 10³(or 1000) and then I have it written in scientific notation as 3.5 X 10-3, which is not actually the formula describing the process to get scientific notation, but rather the formula that reverts it.

Now, when converting a large number- e.g 13,000,000 - into scientific notation, I figured out you have to divide it so as to get a number between 1 and 10, something that can't be achieved - I thought - if the exponentiation is doing successive multiplications.

So I ended up figuring out a negative exponent, to be capable of reverting the scientific notation described on the first example above, which turns a 3.5 back into 0.0035, had to be performing successive divisions.

So, for my second example I have to do 13,000,000 ÷ 107 or 13,000,000 x -7, in either case getting 1.3, so that 13,000,00 becomes 1.3 x 107, again, the resulting formula, i.e scientific notation for this number is not describing the actual process to get the number converted, but rather the formula that reverts this process.

Now, I'm trying but still haven't wrapped my head around this: why negative exponents are performing successive divisions? Why aren't exponentiations always performing multiplications? I don't know if it's only bad memory but I always remembered exponentiation exclusively as an abbreviated form of writing successive multiplications of a number by itself, never as divisions.


2 Answers 2


Note that $$10^{-1}=0.1=\frac {1}{10}$$ Thus $$ 10^{-2} = 10^{-1} \times 10^{-1} = \frac {1}{10}\times \frac {1}{10}= \frac {1}{100}$$

Thus we are still multiplying which means we add powers but powers are negatives.

  • $\begingroup$ Quite interesting, fascinating in fact, because we can represent the process in various manners and still be correct. $\endgroup$ Nov 25, 2018 at 10:22
  • 1
    $\begingroup$ @EzequielBarbosa No, I don't think so. We aren't representing any process in various manners and getting the same answer, we are just doing the same thing and stating mathematical identities. It is factual that $xy^{-1}=x/y$ for any real $x,y$. $\endgroup$
    – YiFan
    Nov 25, 2018 at 12:53

One way to remember this is that a positive exponent means you are inserting factors, while a negative exponent means you are removing factors.

So, $1.3\times 10^{3}=1.3\cdot(10\cdot 10\cdot 10)=1300$, while $1.3\times 10^{-3}=1.3\div (10\cdot 10\cdot 10)=0.0013$.

Addendum: You seem to be confused by the process. It is true that what you have to do to the number to arrive at the correct notation is “opposite” in some sense. Let me show you why.

Suppose you want to write the number 571000 in scientific notation. I think you have no problem coming up with the mantissa (the part out front without the power of $10$). It’s just $5.71$, right? Always a single nonzero digit followed by the decimal point and then the rest of the digits.

Now the thing is to ask, “If I start with the mantissa $5.71$, what do I need to do to it to end up with the original number?” Since this always amounts to shifting the decimal point either right or left, it always means throwing in repeated multipliers (to move the decimal point to the right) or divisors (to move it left) of $10$. In this case, starting with $5.71$, you need to move the decimal point $5$ places to the right to reconstruct the original number, so $$571000=5.71\times 10^5$$

Note that I did this a little differently than you did. You are focusing on how to produce the mantissa (divide $571000$ by $10^5$) and then undoing that by multiplying by $10^5$. The reason they are the same is that $$571000=571000\times(\underbrace{10^{-5}\times 10^5}_1)$$ $$=(571000\times 10^{-5})\times 10^5$$ $$=(5.71)\times 10^5$$

But you can skip this step and just write down the mantissa directly by observation (always of the form $n.nnnnn$ etc) and ask “What else do I need to recover the original number from this mantissa?”

Hope this helps.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.