x^0 = 1 (1 Viewer)

[tommykins]

Can someome provide a proof of this?

Was doing some math and noticed this simple "fact" and wondered how I'd prove it.

Haven't had a go yet, too busy studying but if someone can type up a proof, it'd be awesome thanks.

 

[DJel]

From Maths in Focus textbook;

 

[cwag]

1= aⁿ/aⁿ
1=a^n-n
1=a⁰

 

[cwag]

From Maths in Focus textbook;

haha beat me to it

 

[Gay Captain]

It's pretty simple

can you prove x^m/x^n = x^(m-n) though?

 

[tommykins]

Ahh thanks for that, never thought about it

 

[Slidey]

So what's 0⁰?

Edit: sup not sub

 

[Deadlyned]

It's pretty simple

can you prove x^m/x^n = x^(m-n) though?

x^m/x^n = x^(m-n) is a simple proof by observation... It you have x multiplied by x; m times, and you divide this by x multiplied by x; n times, n lots of x will cancel out from top and bottom, hence leaving only m-n lots of x multiplied by x.

 

[cwag]

It's pretty simple

can you prove x^m/x^n = x^(m-n) though?

x^m/xⁿ
= x*x*x*x*x.. (m times)/ x*x*x*x*x..(n times)
= x*x*x*x....(m-n times)/1
= a^m-n

 

[cwag]

So what's 0₀?

0⁰ = 1 yeah?

 

[Trebla]

Actually 0⁰ = 1 is some contexts and 0⁰ is undefined in others....which is quite interesting actually. You might want to check this out:
http://en.wikipedia.org/wiki/Exponentiation#Zero_to_the_zero_power

 

[Slidey]

0⁰ is like 0!. Except in the case of 0⁰ there are some valid arguments for leaving it undefined in a few cases. From a set-theoretic or combinatorial viewpoint, it equals 1. From a limit/continuity viewpoint, it is undefined.

Likewise,

and

are undefined.

 

[Affinity]

It could be included as an actual number

 

[Affinity]

For example you can declare infinity to be greater than all real numbers
Infinity > all real numbers
Infinity + any number = infinity

Ofcourse you won't have all the properties of real numbers, but this set up is quite useful in some contexts

 

[darkliight]

The equivalence class of unbounded sequences. This then fits in nicely with the standard definition of the real numbers (provided we're careful with the arithmetic).

 

[gurmies]

what's the qualm with defining 1^infinity? Sure infinity itself is not defined, but surely 1 in any power is 1?

 

[Slidey]

For an example of this discrepancy for 1^infinity, on the one hand,
take f(x) = 1 and g(x) = 1/x. Then:

lim f(x)^g(x) = lim 1^(1/x) = lim 1 = 1

On the other hand, if we take f(x) = 1 + x and g(x) = 1/x, then:

lim f(x)^g(x) = lim (1 + x)^(1/x) = e = 2.718281828459... > 1

All as x -> 0.

More here: http://mathforum.org/library/drmath/view/56006.html

 

[shaon0]

If you tried to prove this by using logs it is:
Assume x^0=1
LHS=In(x^0) RHS=In(1)
LHS=0In(x) RHS=In(1)
LHS=0 RHS=0
Therefore, LHS=RHS
Thus, x^0=1

 

[kaz1]

If you tried to prove this by using logs it is:
Assume x^0=1
LHS=In(x^0) RHS=In(1)
LHS=0In(x) RHS=In(1)
LHS=0 RHS=0
Therefore, LHS=RHS
Thus, x^0=1

Multiplying by 0s both sides? I don't think that's allowed.

 

[shaon0]

Multiplying by 0s both sides? I don't think that's allowed.

wat? i didn't do that.