How do you simplify #5^0#?

Question

8104 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-14T04:31:22

Please see below.

While it is easy to understand #5^2# and #5^5#, which are #5xx5# and #5xx5xx5xx5xx5#

it is not easy to understand #5^0#.

So let us try it different way.

What is #5^5-:5^2#?

To answer this let us use division as fraction #5^5/5^2# and this is

#(5xx5xx5xx5xx5)/(5xx5)#

= #(5xx5xx5xxcancel5xxcancel5)/(cancel5xxcancel5)#

= #5^3# and we can express this as #5^((5-2)# i.e.

#5^5-:5^2=5^((5-2)#

In fact similarly we can generalize it to say

#a^m-:a^n=a^m/a^n=a^((m-n)#, which is an identity

What is #a^0# then. Well we can use above identity and can interpret it as

#a^0=a^((m-m)#, but RHS is #a^m-:a^m# or #a^m/a^m# i.e. #1#

Hence for any #a#, we have #a^0=1# and so too for #a=5#

and #5^0=1#