Comments on sum of like powers

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Re: (a+b)^1 = (a+b)

We define powers of a number this way.

For a number b,

 

b^2 = b*b

b^3 = b * b * b

...

b^m = b * b * b .... * b  ( m times)

So we extend the notation to

b^1 = 1 product of b which is just b.

 

b^1 = b for every number b.

Therefore (a+b)^1 = (a+b) for every number (a+b).

Also,  (a+b)^0 = (a+b) multiplied by itself 0 times.

Since 1 is the identity for multiplication, 1 * e = e for every number e,

 

we interpret (a+b) multiplied by itself 0 times to be 1.

 

(a+b)^0 = 1 for all numbers (a+b).

 

Kermit

 

 

 

 

 

posted by Kermit1941 on May 31, 2013 at 9:51 AM | link to this | reply

I'm afraid that I am still pondering (a + b)1 = a + b

posted by TAPS. on May 30, 2013 at 10:53 PM | link to this | reply