Powers of ten addition with negative power I have been stumped by my nephew's algebra homework.
He has a question:
$10^5 + 10^{-3}=$?
The multi-choice answers were:
$10^{-5},10^{15},10^{3},10^{2},10^{5},10^{8}$
(to 1 s.f.)
I thought it was as simple as adding the powers - therefore being $10^2$. Apparently not. Can someone help so I can assist?
 A: The correct answer to the question is $10^5$!!!
Rounded to 1 s.f. (1 significant figure) $10^5 + 10^{-3} = 10000.001$ gives $10^5$.
In school books they write quite often things like
$$10000.001 = 10000 \mbox{ (1 s.f.) }$$
and mean that the number on the right-hand side is the result after rounding the number on the left-hand side to 1 significant figure.
A: You're confusing two identities. The one you're using incorrectly is:
$$10^5\color{red}{\times}10^{-3}=10^{5+(-3)}=10^2\color{red}{\neq}10^5\color{blue}{+}10^{-3}$$
For the right answer, you need to know the values of $10^5$ and $10^{-3}$ individually and then add them. The correct answer is approximately close to the answer you'll get (the problem should ask you to select the closest match after rounding off).
A: $10^5 + 10^{-3} = 10^5 + \frac{1}{10^3} = \frac{10^8+1}{10^3}$.
Clearly the answer matches none of the given choices.
A: Adding two powers of ten never gives a new power of ten. You can see this easily by considering two numbers with a single $1$ (either before or after the decimal point) and otherwise all zeroes. Adding them together either gives a number with two $1$'s and otherwise all zeroes, or a single $2$ and otherwise all zeroes.
A: I will assume that by $10^5+10^{-3}$ you meant $10^5\times 10^{-3}$.

Say we have a value $x$ and we want to raise it to a power $n$. This means that we multiply $x$ by itself $n$ times. $$x^n = \underbrace{x\cdot x\cdot x\cdot\ldots\cdot x}_{n\text{ times.}}\tag1$$ It is confusing to most people when we say that $x^0 =1$, because how can we multiply $x$ by itself $0$ times? Here, we have to look at one of the Power Rules.
The first power rule is as follows: $$x^a\cdot x^b = x^{a + b}.\tag2$$ This is provable from $(1)$. Since $x^a$ and $x^b$ are all products of $x$, then when we multiply them together, the number of times $x$ is being multiplied by itself in total is of course $a + b$ times. Therefore, the product of $x^a$ and $x^b$ is always $x^{a+b}$.
So this means that since $n = n + 0$, we get that $x^n = x^n\cdot x^0$. Therefore, $x^0$ must be equal to $1$. $$x^0 = 1.$$

Now, substitute $x = 10$, $a = 5$, and $b = -3$ in $(2)$. You should have, $$\begin{align} 10^5\times 10^{-3} &= 10^{5+(-3)} \\ &= 10^{5-3} \\ &= 10^2.\end{align}$$
