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As a student you might have studied that any number raised to power zero (0) is always one (1), no matter what the number is – whole number, integer, rational number or irrational number.

You might have used the expression $2^{0}$ as 1, while solving mathematical problems. But do you know why $2^{0}$ = 1?

## Why 2 to the Power of 0 is 1?

There are many ways by which we can prove that any number raised to power 0 is 1. Let’s look at some of the ways where you can demonstrate or prove that $a^{0}$ = 1.

### Method 1

Any number divided by itself gives 1, i.e, $a\div a = 1$.

For example $5\div 5 = 1$, $129\div 129 = 1$, $\frac{3}{2}\div\frac{3}{2} = 1$, …

Let’s start with a÷a = 1

$=>\frac {a}{a} = 1 =>\frac {a^{1}}{a^{1}} = 1 $ (Since, $a^{1} = a$)

Now, recall a law of exponent for division $\frac{a^{m}}{a^{n}} = a^{m-n} $

So, $\frac {a^{1}}{a^{1}}$ becomes $a^{1 – 1} = a^{0}$

And hence, $a^{0} = 1$.

### Method 2

Let’s consider a sequence of powers of 3 in descending order

$3^{6} = 729$, $3^{5} = 243$, $3^{4} = 81$, $3^{3} = 27$, $3^{2} = 9$, $3^{1} = 3$

Did you notice a pattern in the above sequence of numbers?

$3^{5} = 243 = 729 \div 3$

$3^{4} = 81 = 243 \div 3$

$3^{3} = 27 = 81 \div 3$

$3^{2} = 9 = 27 \div 3$

$3^{1} = 3 = 9 \div 3$

The pattern is

- Power is decreasing by 1
- The number on right hand side is obtained by dividing by 3

Following the pattern, the next number will be

$3^{1-1} = 3^{0} = 3\div 3 = 1$

So, again we see that $3^{0} = 1$, or in general $a^{0} = 1$.

### Method 3

In this method you can use the multiplication rule of exponents to deduce that $a^{0} = 1$.

According to multiplication rule of exponents $a^{m} \times a^{n} = a^{m + n}$.

For example $a^{3} \times a^{4} = \left( a \times a \times a \right) \times \left(a \times a \times a \times a \right) = \left(a \times a \times a \times a \times a \times a \times a \right) = a^{7}$

Or, $a^{6} \times a^{2} = (a \times a \times a \times a \times a \times a) \times (a \times a) = (a \times a \times a \times a \times a \times a \times a \times a) = a^{8}$

Can you notice the shortcut here?

$\left(a × a × a × a × a × a … m \ times \right) × \left(a × a … n \ times \right) = \left(a × a × a × a × a × a × a × a … \left(m + n \right) \ times \right)$

i.e., $a^{m} \times a^{n} = a^{m+n}$

Now, consider any of the following two cases:

**Case 1: $m = 0$**

When $m = 0, a^{m} \times a^{n} = a^{m + n}$ becomes $a^{0} \times a^{n} = a^{0 + n} => a^{0} \times a^{n} = a^{n} $

$=> 1 \times a^{n} = a^{n} $ (Since, the product of a number multiplied by another number will be the same **only when the other number is 1**).

$=>a^{0} = 1$

**Case 2: $n = 0$**

When $n = 0, a^{m} \times a^{n} = a^{m + n} $ becomes $a^{m} \times a^{0} = a^{m + 0} => a^{m} \times a^{0} = a^{m}$

$=>a^{m} \times 1 = a^{m}$ (Since, the product of a number multiplied by another number will be the same **only when the other number is 1**).

$=>a^{0} = 1$.

## Practice Problems

- Find the value of
- $\left(2^{5} \right)^{0}$
- $\left(2^{0} \right)^{5}$
- $\left(3^{-6} \right)^{0}$
- $\left(3^{0} \right)^{-6}$
- $\left( \frac {1}{5^{0}}\right)^{0}$

- Simplify
- $2^{1 – 0} \times 2^{0 – 1} \times 2^{1} \times 2^{0} $
- $\frac{5^{2} \times 5^{-2} \times 5^{2 + 2} \times 5^0}{3^{-3} \times 3^{-3 + 3} \times 3^{0} \times 3^{3}} $

## Recommended Reading:

- How to Avoid Silly Mistakes in Maths
- Online Maths Competition in 2022
- Super Interesting Examples of Maths in Real-World

Image Credit: Math teacher vector created by storyset – www.freepik.com