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#11 - ingabenwetrust (09/13/2013) [-]
Can someone on this site explain to me why n^0=1? I've never understood it
#95 to #11 - revisandbutthead (09/14/2013) [-]
Here's the best, simplest way I've heard algebraically.
#79 to #11 - anonymous (09/14/2013) [-]
2^3= 2^2^2= 8
2^2= 2*2 = 4
2^1= 2 = 2
If you want to pass to 2^3 to 2^2, you have to divide by 2 :
8/2 = 4 = 2^2
4/2 = 2 = 2^1
So for 2^0, it's the same :
2^1/2 = 2/2 = 1
I post as anonymous, if i said **** nobody's know it's me.
#64 to #11 - eltupi (09/14/2013) [-]
(x^4)/(x^3) = x^(4-3) = x^1 -- subtract exponent.
Thus:

(x^5)/(x^5)=x^(5-5)=x^0=1 (because z/z = 1)
User avatar #56 to #11 - Mickeyboi (09/14/2013) [-]
Basically it's like this:
If you divide one number into another you subtract the exponents as such; (x^5) / (x^2) = x^3.
Also, every number with no exponent written is understood to have a '1' as the exponent. So something like (x^1) / (x^1) = x^0 = x / x = 1
#51 to #11 - neurofuzzy (09/14/2013) [-]
to provide an alternative to the answer that emrakul gave, you'd have to give a definition before I could tell you "why" n^0=1. If your definition was "x^n is x times itself, n times" then you only have x^n defined for positive integers (so, not zero). If your definition was "x^n is x times 1, n times" then you have it defined for n=zero.

I say that it's an issue of definition because, with certain definitions, x^0 could be left undefined! So, emrakul's answer uses this big and complicated definition, that isn't necessarily needed. It's "because of limits" but only when you start out with that definition. "The true definition" that I'd give a mathematician would define the exponential function in terms of limits, so emrakul's answer would be taking the limit of a limit (or by using continuity, I suppose). Gross.

Eh, but maybe emrakul's answer is satisfactory, because at the heart of it, exponents *are* about integer powers and integer roots. But I'm just saying, I wouldn't overcomplicate it, and I'd just say "x^n is x times 1, n times". Of course that's not "the" definition. You might ask, "what if n is negative, what if n is rational, what if n is irrational?!". But "the" definition is too much machinery and doesn't give intuition w/o a bunch of analysis, and it all builds off the initial definition anyways.
#39 to #11 - anonymous (09/14/2013) [-]
n^2 = n * n.
n^1 = n
n^0 = n / n = 1
n^-1 = 1 / n
n^-2 = 1 / (n * n) = 1 / n^2
See the pattern? That's all it is.
User avatar #30 to #11 - bluemoonz (09/14/2013) [-]
anything to the ****** power is 1
User avatar #29 to #11 - Hreidmar (09/14/2013) [-]
Can I just say I love what you've started? Seeing so many numbers and symbols in a single thread gives me a warn fuzzy feeling inside. I really like knowledge.
User avatar #27 to #11 - tomwh (09/14/2013) [-]
3^3 = 3*3*3 = 27
3^2 = 3*3 = 9
3^1 = 3

On the other side, ignoring 3^0

3^-1 = 3/(3^2) = 3/9 = 1/3
3^-2 = 3/(3^3) = 3/27 = 1/9

So:

3^0 = 3/3 = 1

Hopefully that's simpler than using limits and 'n' and stuff.
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#22 to #11 - imadeanaccforthis **User deleted account** has deleted their comment [-]
#20 to #11 - mjoy (09/13/2013) [-]
It's simple, actually: the exponent is how often you multiply 1 with n (if its positive) or divide 1 by n (if it's negative).
So n^3 = 1 * n * n * n and n^-2 = 1 / n / n.
And that means that n^0 = 1.
#19 to #11 - anonymous (09/13/2013) [-]
Two ways you can think about this:

From the rules for exponents we know:
a^b / a^c = a^(b - c)

So
a^0 = a^(d - d) = a^d / a^d and anything divided by itself equals 1, so
a^0 = 1
#17 to #11 - anonymous (09/13/2013) [-]
3^3 = 27
3^2 = 27 / 3
3^1 = (27 / 3) / 3 or 3^1 = 27 / 9
3^0 = (27 / 9) / 3 or 3^0 = 27 / 27
so 3^0 = 1
this works with any numbers you choose
#13 to #11 - emrakul (09/13/2013) [-]
Basically, limits. The graph of y=(n^1)x is a straight line, while the graph of y=(n^1/2)x is a curve that levels off as it gets larger. All exponential functions equal 1 at x=1, and as the exponent gets smaller the curve of the function levels off even more. 10^1/2 is greater than 10^1/4, which is greater than 10^1/8, and so on. As the exponent in the function y=(n^e)x gets smaller, it gets closer and closer to a horizontal line, still touching the point (1,1). So while it might not be intuitively possible to calculate n^0, it's pretty easy to see that lim(e->0) n^e=0.

Sorry for the mathiness.
#91 to #13 - anonymous (09/14/2013) [-]
Cba to log in to post a screen cap, but atm your comment has 42 likes. Felt it was fitting for the picture you uploaded. carry on with your day now good sir .
#75 to #13 - crazyolitis (09/14/2013) [-]
What's going on.
What's going on.
User avatar #63 to #13 - chrismamaril (09/14/2013) [-]
i didn't read your comment but with the amount of thumbs you get i will assume that you are right. oh, god. i'd love to have people like you in my math class. I suck at it we(all of my class mates) all do..
#47 to #13 - anonymous (09/14/2013) [-]
User avatar #41 to #13 - captainrattrap (09/14/2013) [-]
There's something about that comment and the picture that makes me really like that **** . Idk.
User avatar #21 to #13 - itsthatguyagain (09/13/2013) [-]
Thanks for that picture, I love Hitch Hiker's Guide.
User avatar #14 to #13 - emrakul (09/13/2013) [-]
edit: I is confus. Replace all n's with x's in my equations and ignore the x's I wrote.
User avatar #16 to #14 - cosminb (09/13/2013) [-]
Was just about to check with my algebra book when I saw this comment.
Good job, sir!
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