why is x^0 = 1

ryanairzer

and why is (-1)x(-1)=1

henbane

why is x^0 = 1

http://mathworld.wolfram.com/Power.html

MrAbc

why is (-1)x(-1)=1

Wikipedia:
http://en.wikipedia.org/wiki/Negative_and_non-negative_numbers

(-1) (-1)

= ((-1) (-1)) - 2 + 2

= ((-1) (-1)) + ((-1) x 2) + 2

= (-1) x ((-1) + 2) + 2

= (-1) x (1) + 2

= (-1) + 2

= 1

IconoclastCynic

ryanairzer wrote:

and why is (-1)x(-1)=1

I'm not going to try to prove the equations because I assume you're just trying to justify it in your mind as they seem fairly bizarre and non-intuitive.

For x^0 = 1, start off with x^a = b.
When x > 0, b is a relatively large number. This is very easy to verify.
If you take x to be 2 and b to be 2, you get 2^2 = 2*2 = 4, which is twice the magnitude of x.
Now if you take x to be 2 and a to be 3, you get 2^3 = 2*2*2 = 4*2 = 8 - four times as large as x.
For obvious reasons, they say that x is growing 'exponentially'.
Now, let's go in the opposite direction.
Suppose x<0, if we say that x is 2 and a is -1 we get 2^(-1) = 1/(2^1) = 1/2Or if we call a -2, we get 2^(-2) = 1/(2^2) = 1/4.
As you can see, as a gets farther and farther from 0 in the negative direction, b gets closer to 0. And if a get's closer to 0 from the negative direction, b gets closer and closer to 1.
This is true from the other side as well, but you end up dealing with roots, so the numbers get increasingly difficult to deal with.
I suppose if you want to write it mathematically, you'd say:
lim(a->0)[x^a] = 0.

The other one, (-1)*(-1) is much easier to explain.
Suppose you have a number x. x can be either negative or positive, there is no restraint on the sign of x. If you multiply it by -1, you get -x. All you have done is change the sign. Supposing x is 2, -1*x = -x = -2. It is clear at this point that multiplying by -1 is simply a sign modulator. Thus, if you call x -2,
-1*x = -x = -(-2) = 2. All you have done is change the sign of -2. Since we only have two signs (negative and positive) if you multiply a number by -1 twice you don't change the sign at all yielding (-1)*(-1) = 1. Another way of looking at that is calling x -1. then you have (-1)*x = -x, which just changes the sign. And since your initial sign of x is negative, you simply change it to a positive.

Cunning

well x^1 is x obviously
and x^2 is x^2 obvoiusly

AND x^5/x^3 =x^2 OR x^(5-3) OR xxxxx/xxx =xx

so (x^n)/(x^n) is 1 obviously, but is also x^(n-n) =x^0 =1

TA DA

ryanairzer

Thanks y'all.

Delphi91

Consider the following:

(x^a)/(x^b) = x^(a-b)

But if a=b then you get x^0.

And the fraction (x^a)/(x^b) has the same numerator and denominator which means that the value of the fraction is 1.

Therefore x^0 = 1!

shanethemofo

Delphi91 wrote: »

Consider the following:

(x^a)/(x^b) = x^(a-b)

But if a=b then you get x^0.

And the fraction (x^a)/(x^b) has the same numerator and denominator which means that the value of the fraction is 1.

Therefore x^0 = 1!

But what when (x^a)/(x^b) = (x^0)/(x^0)



Endless circle methinks :P

MrAbc

shanethemofo wrote: »

But what when (x^a)/(x^b) = (x^0)/(x^0)

then (x^a)/(x^b) = 1

simply implies a=b

just like (x^a)/(x^b) = log[(a^b)]/log[(a^b)] => a=b ;-)

jackdaw

For a real world explanation of why (-)(-) = + .

imagine this

if i make €20 (+20) a day, in 5 days (+5) i will have made +5 * +20 = €100 Euro

If i am losing €20 a day (-20) then 5 days AGO (-5) i will have -5 * -20 = €100
more (+) of what I have now..

vallo

jackdaw wrote: »

For a real world explanation of why (-)(-) = + .

or .... I'm not not going to explain it to you.

..of course I would then have to proceed to explain it to you, but I think the others have done a persuasive job.
I really like the x^0 explanations. Very cunning indeed.