Exponents

Ontario 09 Academic (MPM1D)

Zero power with integer bases

Lesson

In the lesson about the division law we learned that we can simplify the division of terms that have like bases by finding the difference between the exponents (or powers).

So what happens when we perform the subtraction and we are left with a power of $0$0? For example,

$4^1\div4^1$41÷41 | $=$= | $4^{1-1}$41−1 |

$=$= | $4^0$40 |

To think about what value we can assign to the term $4^0$40, let's write this division problem as the fraction $\frac{4^1}{4^1}$4141. Since the numerator and denominator are the same, the fraction simplifies to $1$1. Notice that this will also be the case with $\frac{4^2}{4^2}$4242 or any expression where we are dividing like bases whose powers are the same.

So the result we arrive at by using exponent laws is $4^0$40, and the result we arrive at by simplifying fractions is $1$1. This must mean that $4^0=1$40=1.

There is nothing special about the number $4$4, so we can extend this observation to any other base. This result is summarised by the zero power law.

The zero power law

For any base number $a$`a`,

$a^0=1$`a`0=1

This says that taking the zeroth power of any number will always result in $1$1.

Find the value of $23^0$230.

**Think**: We are looking for the number that is equivalent to the term $23^0$230. Notice that the base is $23$23 and the power is $0$0.

**Do**: Using the zero power law, we have $23^0=1$230=1.

**Reflect**: In this case the base was $23$23, but the result is the same no matter what base we have. Here are some more expressions that are also equal to $1$1: $23456^0$234560, $9.75^0$9.750, $\left(\frac{3}{7}\right)^0$(37)0, $\pi^0$π0.

It does not take long to get used to the zero power law for individual numbers. But what about more complicated expressions? Remember that when an expression is enclosed in brackets, then the power is applied to everything inside the brackets. This is no different when the power is $0$0.

Find the value of $\left(2+6\times8\right)^0$(2+6×8)0.

**Think**: This expression is in the form $a^0$`a`0, where the base is $\left(2+6\times8\right)$(2+6×8) and the power is $0$0. This means that we can use the zero power law.

**Do**: Using the zero power law, we have $\left(2+6\times8\right)^0=1$(2+6×8)0=1.

**Reflect**: By identifying that the power is acting on the whole expression within the brackets, we could short-cut our way to the solution without needing to perform any multiplication or addition. Here is a longer path to the same solution that shows the steps we skipped.

$\left(2+6\times8\right)^0$(2+6×8)0 | $=$= | $\left(2+48\right)^0$(2+48)0 |

$=$= | $50^0$500 | |

$=$= | $1$1 |

Evaluate $741^0$7410.

Evaluate $65^0+81^0$650+810.

Evaluate $72^0+2^3$720+23.

Derive, through the investigation and examination of patterns, the exponent rules for multiplying and dividing monomials, and apply these rules in expressions involving one and two variables with positive exponents