This expression … | means … |
---|---|

… | … |

4 · a, |
a + a + a + a; |

3 · a, |
a + a + a; |

2 · a, | a + a; |

1 · a, | a. |

Can it go any further downwards?
That is, can we extend it
to describe multiplication by any whole number (including zero)
or even any integer (including the negative integers)?
The pattern that I've written so far really doesn't continue;
at each stage, we remove one more addition of *a*,
one more ‘+ *a*’ disappears,
but the last step, 1 · *a*, doesn't really have any addition,
and there is no more plus sign to remove.
After all, *a* added to itself *one* time
isn't really *a* added to anything at all!

However, I can tweak this table slightly
to allow the pattern to continue one more step.
We know that 0 · *a* is supposed to be 0,
and you might think that this is what you get
when you remove *everything* from the last line of the table.
This really only works, however, if you think that a calculation begins with 0.
In the case of repeated addition, that really is true,
so let's make it explicit:
to multiply *a* by (say) 3,
start with 0, then add *a* to *that* 3 times:
0 + *a* + *a* + *a*
instead of simply *a* + *a* + *a*.
Then the table looks like this:

This expression … | means … |
---|---|

… | … |

4 · a, |
0 + a + a +
a + a; |

3 · a, |
0 + a + a + a; |

2 · a, |
0 + a + a; |

1 · a, | 0 + a; |

0 · a, | 0. |

Can we continue this to multiplication by negative integers?
Not exactly, but we can start a similar, reverse pattern
by changing *addition* to the inverse operation of *subtraction*.
The final table looks like this:

This expression … | means … | or … for short |
---|---|---|

… | … | … |

4 · a, |
0 + a + a + a + a, |
a + a + a + a; |

3 · a, |
0 + a + a + a, |
a + a + a; |

2 · a, |
0 + a + a, |
a + a; |

1 · a, |
0 + a, | a; |

0 · a, | 0, | 0; |

(−1) · a, |
0 − a, | −a; |

(−2) · a, |
0 − a − a, |
−a − a; |

(−3) · a, |
0 − a − a − a, |
−a − a − a; |

(−4) · a, |
0 − a −
a − a − a, |
−a − a −
a − a; |

… | … | … |

I've put a new column to the right, giving the simplified expression that you would usually use. (Why write in all those zeroes if you don't need them?) But the middle column is the one that really shows the pattern: start with 0, then add or subtract as necessary.

As the first example,
consider (2/7)^{3},
which means (2/7) · (2/7) · (2/7),
which works out to 8/343.
Again, there's really nothing special about 2/7;
in general, the rule is that *a*^{3}
means *a* · *a* · *a*.
Here, *a* is called the **base**
and 3 is the **exponent**;
the result *a*^{3} is the **power**,
or in full ‘the 3rd power of *a*’.
You can also read this as ‘*a* raised to the 3rd power’,
or simply ‘*a* to the 3rd’,
or even ‘*a* cubed’
(a nickname that comes from geometric applications).

Here is the table (analogous to the first table above) for raising numbers to a natural exponent:

This expression … | means … |
---|---|

… | … |

a^{4}, |
a · a ·
a · a; |

a^{3}, |
a · a · a; |

a^{2}, |
a · a; |

a^{1}, | a. |

What is *a*^{0} then?
In other words, how do we extend this table one more line
to make the analogue of the second table above?
You might think that *a*^{0} is 0,
just like 0 · *a* is 0,
on the grounds that you start calculating at 0.
But in fact that does *not* work for repeated multiplication;
when you multiply, you really start calculating at 1!

To convince yourself of this, just imagine that you replace 1 with 0 in the table below. Then all of the results would simply be 0, which is not correct; only if you use 1 instead can you get the same answers as before.

This expression … | means … |
---|---|

… | … |

a^{4}, |
1 · a · a ·
a · a; |

a^{3}, |
1 · a ·
a · a; |

a^{2}, |
1 · a · a; |

a^{1}, | 1 · a; |

a^{0}, | 1. |

What is the analogue for multiplication of subtraction? It is division, although there is an extra complication: you cannot divide by 0. Therefore, you cannot raise 0 to a negative exponent.

Here is the final table, analogous to the final table in the previous section:

This expression … | means … | or … for short; |
---|---|---|

… | … | … |

a^{4}, |
1 · a ·
a · a · a, |
a · a ·
a · a; |

a^{3}, |
1 · a ·
a · a, |
a · a · a; |

a^{2}, |
1 · a · a, |
a · a; |

a^{1}, |
1 · a, | a; |

a^{0}, | 1, | 1; |

a^{−1}, |
1/a, | 1/a; |

a^{−2}, | 1/a/a, |
1/(a · a); |

a^{−3}, | 1/a/a/a, |
1/(a · a · a); |

a^{−4}, |
1/a/a/a/a, |
1/(a · a ·
a · a); |

… | … | … |

To form the usual (short) expressions in the right column,
I can use rules for dividing fractions.
For example, 1/*a*/*a*, or 1/*a* divided by *a*,
means 1/*a* multiplied by the reciprocal of *a*,
or (1/*a*) · (1/*a*),
which works out to (1 · 1)/(*a* · *a*),
or 1/(*a* · *a*).

Sometimes people say that 0^{0} is also undefined,
but this isn't really correct.
The definition works perfectly well to produce 0^{0} = 1.
Of course, you can always make an exception in a definition
if you want something to be undefined,
and in fact mathematicians argued about this for almost 200 years.
But in the last 50 years or so,
people have begun to actually *use* 0^{0},
and they find that 0^{0} has to be defined as 1
for a bunch of other things
(mostly in the mathematical field of combinatorics)
to work.

Incidentally, this is how such arguments usually get settled:
peeople go back and forth for a while,
getting nowhere as long as it's purely theoretical;
but once somebody finds a way to *use* the ideas,
then it's obvious what the right answer has to be.
Sometimes it takes the textbooks a while to catch up, however;
the textbook for this course still says that 0^{0} is undefined,
but in doing so it is behind the times!

Notice that the sign (positive or negative) of the exponenet
has *nothing* to do with the sign of the power.
For example, 2^{2} = 2 · 2 = 4,
while 2^{−2} = 1/(2 · 2) = 1/4,
and these are both positive.
This is because a negative exponent
simply tells you to divide instead of multiplying,
which has nothing to do with whether your result is negative.

On the contrary, if the base is negative,
then the sign of the power
depends on the *parity* (even or odd) of the exponent.
For example,
(−2)^{2} = (−2) · (−2) = 4,
while (−2)^{3} =
(−2) · (−2) · (−2) =
(−2) · 4 = −8.
This is because multiplying (or dividing) by a negative number
always switches the sign,
so as you raise a negative number to a power,
you keep going back and forth between negative and positive
until the sign of the final answer
depends on whether the number of times you flip is even or odd.

When you raise a fraction to a power,
you can do the numerator and denominator separately.
For example, (2/3)^{2} = (2/3) · (2/3) =
(2 · 2)/(3 · 3) = 4/9,
and you get the same result if you do (2^{2})/(3^{2}).
Similarly, my original example (2/7)^{3}
can be worked out simply
as (2^{3})/(7^{3}) =
(2 · 2 · 2)/(7 · 7 · 7) =
8/343.
This is because multiplying fractions
amounts to simply multiplying the numerators and denominators separately.
Furthermore, when repeatedly multiplying the same numbers
(as we do in exponentiation),
you'll never have to reduce the fraction
as long as it was reduced originally.

When raising a fraction to a negative exponent,
the easy thing to do is to take the reciprocal first.
For example, (2/3)^{−2} = (3/2)^{2},
since 3/2 is the reciprocal of 2/3,
so (2/3)^{−2} is simply 9/4.
This is because raising to a negative exponent means division,
which is simply multiplication by the reciprocal.
As another example, (1/2)^{−3} = 2^{3},
because 2 is the reciprocal of 1/2,
so (1/2)^{−3} works out to 8.
Conversely, 2^{−3} is 1/8.

Also, you should be careful with the order of operations;
unless grouping symbols (like parenthese) get in the way,
exponentiaton always comes first.
For example, −4^{2}
means −(4 · 4), which is −16,
while (−4)^{2}
means (−4) · (−4), which is 16.
Also, 2/3^{2} means 2/(3 · 3), which is 2/9,
while (2/3)^{2} means (2/3) · (2/3), which is 4/9.

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This web page was written between 2007 and 2013 by Toby Bartels, last edited on 2013 October 10. Toby reserves no legal rights to it.

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