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vicky877 · 28th November 2007, 06:41 AM

hi what u think about that is it true or wrong
comment..
(x)^0=1
justify yourself..

Pommie · 28th November 2007, 06:50 AM

1000, 100, 10, 1, 0.1, 0.01, 0.001

or,

16,8,4,2,1,½,¼

Notice a pattern.

Mike.

jpanhalt · 28th November 2007, 09:11 PM

x^n/x^n = 1

Which is to say, x^(n-n) = 1
X^0 = 1
QED

John

ljcox · 1st December 2007, 09:53 PM

The infinite series for (1 + x)^n = 1 + nx + n(n-1) x/2 + n(n-1)(n-2) x^2/6 + ...

Note that n is in all terms except the first.

Therefore if n = 0, then (1 + x)^n = 1.

QED

tkbits · 4th December 2007, 01:21 AM

0 ^ 0 is undefined.

For n > 0, 0 ^ n = 0.
For x not 0, x ^ 0 = 1.

vicky877 · 4th December 2007, 08:31 AM

Quote:

Originally Posted by tkbits

0 ^ 0 is undefined.

For n > 0, 0 ^ n = 0.
For x not 0, x ^ 0 = 1.

MAN U ARE REALLY VERY CONCEPTUALLY ..

ljcox · 4th December 2007, 10:25 PM

Quote:

Originally Posted by tkbits

0 ^ 0 is undefined.

For n > 0, 0 ^ n = 0.
For x not 0, x ^ 0 = 1.

I would have thought so, but if you set x = -1 in the series I posted above, 0^0 = 1.

???

wschroeder · 6th December 2007, 01:34 AM

Division by 0 is undefined for real numbers. If we use the example 0^1/0^1 = 0^(n-n) you have a problem to declare or prove 0^0 = 1.

ljcox · 6th December 2007, 08:43 AM

Quote:

Originally Posted by wschroeder

Division by 0 is undefined for real numbers. If we use the example 0^1/0^1 = 0^(n-n) you have a problem to declare or prove 0^0 = 1.

I did not use the 0^1/0^1 = 0^(n-n) proof. I used the series posted above which does not have any division by zero.

However, I can't see how 0^0 can = 1.

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28th November 2007, 06:41 AM	(permalink)
vicky877	maths hi what u think about that is it true or wrong comment.. (x)^0=1 justify yourself.. __________________ vicky,s

28th November 2007, 06:50 AM	(permalink)
Pommie	1000, 100, 10, 1, 0.1, 0.01, 0.001 or, 16,8,4,2,1,½,¼ Notice a pattern. Mike.

28th November 2007, 09:11 PM	(permalink)
jpanhalt	x^n/x^n = 1 Which is to say, x^(n-n) = 1 X^0 = 1 QED John

1st December 2007, 09:53 PM	(permalink)
ljcox	The infinite series for (1 + x)^n = 1 + nx + n(n-1) x/2 + n(n-1)(n-2) x^2/6 + ... Note that n is in all terms except the first. Therefore if n = 0, then (1 + x)^n = 1. QED __________________ Len

4th December 2007, 01:21 AM	(permalink)
tkbits	0 ^ 0 is undefined. For n > 0, 0 ^ n = 0. For x not 0, x ^ 0 = 1.

6th December 2007, 01:34 AM	(permalink)
wschroeder	Division by 0 is undefined for real numbers. If we use the example 0^1/0^1 = 0^(n-n) you have a problem to declare or prove 0^0 = 1. Last edited by wschroeder; 6th December 2007 at 01:37 AM.