0^0 = 1

fusionskd · 20-10-2007: 20th October 2007 14:07

It is implied in mathematics that;
0^0 = 1

Can anybody explain why this is, whats the proof behind this?
Doesn't the statement also imply that;
n/n = n^0

when n=0

??????????

73337 · 20-10-2007: 20th October 2007 14:08

LHS does not exist, as such. [I believe

]

Airel · 20-10-2007: 20th October 2007 14:12

So I see you divided by zero.

wait.

ohSHI-

fusionskd · 20-10-2007: 20th October 2007 14:15

On google it says:
http://www.google.co.uk/search?hl=en&q=0%5E+0&meta=

0^0 = 1

Also I have read on other mathematical sources that 0^0 does equal 1 for the binomial theorem to work and other reasons. Can somebody explain this please.

fusionskd · 20-10-2007: 20th October 2007 14:18

Also here is why it is defined as 1:
http://www.math.hmc.edu/funfacts/ffiles/10005.3-5.shtml

if this is the case, doesnt it imply that 0/0=1

?

73337 · 20-10-2007: 20th October 2007 14:20

Well it does say it is undefined.

fusionskd · 20-10-2007: 20th October 2007 14:23

But there are other mathematical sources that have defined it as '1', the question i'm asking is that doesnt that also imply
0/0 = 1

heres another:
http://home.att.net/~numericana/answ...bra.htm#zeroth

Gaz031 · 20-10-2007: 20th October 2007 14:29

This is false for n=0, the symbol p/q (whether p,q be rational, real or complex) has no meaning for q=0

sevengreenmice · 20-10-2007: 20th October 2007 14:35

0/0 = 0^0 = 1
but in the special case of divisions by zero, 0/0 or n/0 = ∞

The Bachelor · 20-10-2007: 20th October 2007 14:37

From what I gather, it is very convenient to think "anything to the power of 0 is 1" - then you don't have to worry about special cases. But remember, it is technically an indeterminate form, so there are undoubtedly many limits of functions that will approach 0^0 at some certain point, but the value of the limit may be something different.

(That's my take on it anyway)

http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

Dadeyemi · 20-10-2007: 20th October 2007 16:32

someone tried to do this but just made up a new number nullity which is the answer to the question. i would say that there is no solution i.e. 0^0 does not exist/is not a number.

El Matematico · 20-10-2007: 20th October 2007 16:33

Originally Posted by fusionskd

It is implied in mathematics that;
0^0 = 1

Can anybody explain why this is, whats the proof behind this?
Doesn't the statement also imply that;
n/n = n^0

when n=0

??????????

0^0 is sometimes stated to be 1, as 0^0 is the number of maps from the empty set to itself, which is 1. Vut it is sometimes stated that 0^0 = 0

datr · 20-10-2007: 20th October 2007 17:58

I was shown this at my Cambridge interview.

y=x^x

$\ln y = x\ln x$

$u = \frac{1}{x}$

$\ln y = -\frac{\ln u}{u}$

As $x \to 0$ , $u \to \infty$ and $\frac{\ln u}{u} \to 0$

$\ln y = 0$

y = 1

fusionskd · 20-10-2007: 20th October 2007 18:19

Originally Posted by datr

I was shown this at my Cambridge interview.

y=x^x

$\ln y = x\ln x$

$u = \frac{1}{x}$

$\ln y = -\frac{\ln u}{u}$

As $x \to 0$ , $u \to \infty$ and $\frac{\ln u}{u} \to 0$

$\ln y = 0$

y = 1

Hmmmm interesting...
Can it not also be shown without the substitution?

y=x^x

$\ln y = x\ln x$

x=0

$\ln y = 0\ln0$

$\ln y\ = 0$

hence...

y = 1

but is

equal to

as

is undefined...hmmmm

It would be like saying (0/0)0 = 0

...

generalebriety · 20-10-2007: 20th October 2007 18:28

Originally Posted by fusionskd

$\ln y = 0\ln0$

Well, yes; the other method strikes me as 'more rigorous', though. It's obvious that x ln x tends to 0 as x tends to 0, but it's still not nice.

mr_jr · 20-10-2007: 20th October 2007 18:32

Originally Posted by fusionskd

It is implied in mathematics that;
0^0 = 1

Can anybody explain why this is, whats the proof behind this?
Doesn't the statement also imply that;
n/n = n^0

when n=0

??????????

You've said
n/n = n^0

implies
$\larrow0/0 = 0^0$

then that $\frac{0}{0}$ makes sense, which of course it doesn't.

It could just be defined as so, in the same way that

is defined to be 1. Its just definition...

fusionskd · 20-10-2007: 20th October 2007 18:38

The definition for

actually works and is easy to understand via the proof.

We know that;
n(n-1)! = n!

hence...

Zii · 20-10-2007: 20th October 2007 18:54

When

it could be either. Or any number you like really, as you can't divide any number by zero, or can you take the log of zero. It's daft to think otherwise. Here's a clear illustration...

$0 \times 5 = 7 \times 0$

$\frac{0}{0} = \frac{7}{5}$

Or... any *******ing number you like! Also... if $3 \times 5 = 15$ then $\frac{15}{5} = 3$ . Assume you can do the same thing with zero. $0 \times 19 = 0$ .

19/0 =

...?

? But $0 \times 0$ is zero...

Thus NEVER DIVIDE BY ZERO. Also no logging! And no 0^0

either. It's all just silly. Hence why mathematicians define it all as undefined.

And for ANYONE who thinks $\frac{x}{0} = \infty$ you too are also silly. Silly, silly silly.

madima · 20-10-2007: 20th October 2007 20:06

Originally Posted by fusionskd

Also here is why it is defined as 1:
http://www.math.hmc.edu/funfacts/ffiles/10005.3-5.shtml

if this is the case, doesnt it imply that 0/0=1

?

if 1/0 is infinity and 0/0 is identical to 0(1/0) then this is the same as 0 multiplied by infinity, which is 0 i guess... i just realised. this is completely irrelivent

generalebriety · 20-10-2007: 20th October 2007 20:10

Originally Posted by madima

if 1/0 is infinity and 0/0 is identical to 0(1/0) then this is the same as 0 multiplied by infinity, which is 0 i guess... i just realised. this is completely irrelivent

And wrong.

1/0 has no meaning, regardless of whether you follow that up by multiplying by 0 too. The 0s don't cancel because "1/0" is meaningless. It's like "1/+" or "1/tree". Division is not defined under those circumstances.

On the other hand, fractions can approach a limit as their numerators and denominators approach zero (depending on 'how fast' they're going to zero and their relative sizes).

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0^0 = 1

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