# How do you write a convincing argument to show why 3^0=1 using the following pattern: 3^5=243, 3^4=81, 3^3=27, 3^2=9?

##### 2 Answers
Sep 26, 2017

Refer to the Explanation.

#### Explanation:

${3}^{5} , {3}^{4} , {3}^{3} , {3}^{2} , {3}^{1} , {3}^{0} , \ldots$

$= 243 , 81 = \frac{243}{3} , 27 = \frac{81}{3} , 9 = \frac{27}{3} , 3 = \frac{9}{3} , 1 = \frac{3}{3.}$

$\therefore {3}^{0} = 1.$

Sep 26, 2017

See argument below

#### Explanation:

The most convincing argument I can think of off the bat is simply:

${x}^{0} = 1 \forall x \in \mathbb{R}$

However, I guess that was not the argument being sought here!

Here we have a sequence: ${a}_{n} = {3}^{n}$ for $n = 5 \to 2$

We can extend this to: ${a}_{1} = {3}^{1} = 3$

Clearly: ${a}_{n - 1} / {a}_{n} = \frac{1}{3} \to {a}_{n - 1} = {a}_{n} / 3$

Now consider ${a}_{0} = {a}_{1} / 3 = \frac{3}{3} = 1$

Since ${a}_{0} \equiv {3}^{0}$ the proposition is proved.