Power of One

[sologuitar]

Let a = any integer 

Why does a^(1) = a?

 

 

[Eise]

See here:

 

[sologuitar]

14 hours ago, Eise said:

See here:

 

Sorry but I don't understand your descending powers in the denominator for each fraction.

For example, you said (a^3) = (a^4)/(a•a^2). 

Can you further explain your work here? 

[Eise]

5 hours ago, sologuitar said:

Sorry but I don't understand your descending powers in the denominator for each fraction.

For example, you said (a^3) = (a^4)/(a•a^2). 

Can you further explain your work here? 

No, I said:

• a^3 = (a^4)/a

There are several other ways to see it, but they all are variations of the same theme:

• a^4 = a x a x a x a
• a^3 = a^4/a = (a x a x a x a)/a = a x a x a
• a^2 = a^3/a = (a x a x a)/a = a x a
• a^1 = a^2/a = (a x a)/a = a
• a^0 = a^1/a =  a/a = 1

In short, the obvious rule is that with division of powers, you subtract the powers:

(a^n)/a^m=a^(n - m).

So when n = m:

(a^n)/(a^n) = a^(n - n) = a^0. 

But dividing two equal numbers always gives 1. (Except 0^0, which can not be defined.)

 

[sologuitar]

3 hours ago, Eise said:

No, I said:

• a^3 = (a^4)/a

There are several other ways to see it, but they all are variations of the same theme:

• a^4 = a x a x a x a
• a^3 = a^4/a = (a x a x a x a)/a = a x a x a
• a^2 = a^3/a = (a x a x a)/a = a x a
• a^1 = a^2/a = (a x a)/a = a
• a^0 = a^1/a =  a/a = 1

In short, the obvious rule is that with division of powers, you subtract the powers:

(a^n)/a^m=a^(n - m).

So when n = m:

(a^n)/(a^n) = a^(n - n) = a^0. 

But dividing two equal numbers always gives 1. (Except 0^0, which can not be defined.)

 

Thank you. A much better reply.