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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?

steve
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    Are you sure that the problem was $10^5 + 10^{-3}$, and not $10^5 \cdot 10^{-3}$? Because none of the answers given is correct for the question you wrote. The right answer is 10000.001, which is not $10$ raised to any integer power. – John Hughes Feb 14 '18 at 15:19
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    The answer is none of the above. Are there other instructions you are leaving out (like rounding to the nearest whole number)? – Umberto P. Feb 14 '18 at 15:19
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    None of the answers are correct. – copper.hat Feb 14 '18 at 15:20
  • Sorry yes - there is small print saying 1 signifcant figure – steve Feb 14 '18 at 15:33
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    Go here $\longrightarrow$ https://math.stackexchange.com/questions/2646922/why-is-everything-except-0-to-the-power-of-0-always-1/2646949#2646949 and read the first half of my answer. Then, substitute $x = 10$, $a = 5$, $b = -3$. – Mr Pie Feb 14 '18 at 15:44

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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.

trancelocation
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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).

Gaurang Tandon
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$10^5 + 10^{-3} = 10^5 + \frac{1}{10^3} = \frac{10^8+1}{10^3}$.

Clearly the answer matches none of the given choices.

Shobhit
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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.

Arthur
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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, $$\begin{align} 10^5\times 10^{-3} &= 10^{5+(-3)} \\ &= 10^{5-3} \\ &= 10^2.\end{align}$$

Mr Pie
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