# Why when indices cancelled out leave 1 at top?

This is a really basic question, but I have just got interested in math and learning rules about powers/indices and this confused me a little.

$\dfrac{a^3}{a^7}$ after they cancel out we get $\dfrac{1}{a^4}$

My question is why do we have 1 remain at the top? Why not 0, any simple explanation for this?

• Try it with $a=2$, say. $\frac0{2^4}=\frac0{16}=0$. Is $\frac{2^3}{2^7}=\frac8{128}$ equal to $0$? (That won’t explain why you get $1$, but it does show that you can’t possibly get $0$.) Commented Mar 5, 2013 at 22:06
• The empty product is $1$ that why it is $1$, it is the neutral element of multiplication Commented Mar 5, 2013 at 22:06
• I'm changing the division-algebras tag. It doesn't mean quite what you think it does. Commented Mar 5, 2013 at 22:10

## 5 Answers

It is really $a^{3-3}=a^0=1$...

The reason is because when you have an expression like that, what you're effectively doing is $$\frac{a^3}{a^7} \div \frac{a^3}{a^3} = \frac{a^3\div a^3}{a^7\div a^3}$$ Where $\dfrac{a}{a}=1$, and $\dfrac{a^3}{a^3} = \dfrac{a}{a}*\dfrac{a}{a}*\dfrac{a}{a}$

"Cancelling" is a misleading expression. You're not cancelling anything, that is, just making something disappear. You're dividing the numerator and denominator by the same thing, in this case, $a^3$. In the numerator, you're dividing $a^3$ by $a^3$, giving 1. In the denominator, you're again dividing by $a^3$. Here you wind up with 3 sets of (a/a), which each turn into 1. and four a's that give the $a^4$

• I should look at any new answers posted while I was composing mine, before posting mine... Commented Mar 5, 2013 at 22:17
• Sometimes words do more than pictures...+1 Nice post Commented Mar 5, 2013 at 22:31

$$\frac{a^3}{a^7}=\frac{a\cdot a\cdot a}{a \cdot a \cdot a \cdot a \cdot a\cdot a\cdot a}=\frac{1\cdot a\cdot a\cdot a}{a \cdot a \cdot a \cdot a \cdot a\cdot a\cdot a}=\frac{1\cdot a^3}{a^3\cdot a^4}=\frac{1}{a^4}$$

Because $1$ is the multiplicative identity not $0$

You don't 'cancel' numerator and denominator, you multiply numerator and denominator by the inverse of the common element in numerator and denominator and in $\frac{a^3}{a^7}$, the common part is $a^3$ itself , so when you multiply numerator and denominator by inverse of $a^3$, you get the multiplicative identity($1$ in this case) on top and $a^4$ in denominator.