Why is it that $2\times 0 = 0$, but $2^0= 1$?
I don't understand the reason behind it. If anyone explain it then it will be very appreciated. I know how multiplication, addition, and division work, but this one always puts my mind back to null.
$\endgroup$ 33 Answers
$\begingroup$Because $\;2 \cdot 3 = \color{red}{0} + \underbrace{2 + 2 + 2}_{3\;\text{times}}\;$, while $\;2^3 = \color{red}{1} \cdot \underbrace{2 \cdot 2 \cdot 2}_{3\;\text{times}}\;$.
$\endgroup$ 7 $\begingroup$Alternatively: $2\cdot0=2\cdot(1-1)=2-2=0$; $2^0=2^{1-1}=\frac22=1$.
$\endgroup$ $\begingroup$We know that for all reals $x$ and $y$ and positive $a$ the following is true. $$a^{x-y}=\frac{a^x}{a^y}.$$ Now, let $x=y$.
We obtain $$a^0=1.$$
$\endgroup$
