/sci/ - Science & Math

[Anonymous No.6465300]

n^0 = 1

Why?

[Anonymous No.6465306]

because n^0 * n = n^1 = n*1

[Anonymous No.6465312]

http://bit.l
y/PTUVIw

Next.

[Anonymous No.6465326]

>>6465300
Do you even logic? to go from n^2 --> n^1 you divide by n, thus you divide n^1 by n to get n^0

(n^1)/(n)=n^0 --> n/n=n^0 --> 1=n^0

[Anonymous No.6465333]

Think of this way,
n^4 = 1*n*n*n*n
n^3 = 1*n*n*n
n^2 = 1*n*n
n^1 = 1*n
n^0 = 1
n^-1 = 1*(1/n)

There are other ways of looking at it though, it's mostly defined this way because it is convenient.

[Anonymous No.6465334]

>>6465300
The only one answer:

n ^ 0 = (n ^ 1) * (n ^ -1) = n / n = 1

[Anonymous No.6465345]

It's quite simple anon. You see n^x is a group homomorphism from (Z,+) to (R, *) and as a morphism it maps the additive identity 0 to the multiplicative identity 1.

[Anonymous No.6465356]

>>6465345
No, that's the wrong order. You define n^x, then you explore its properties. Being a group homomorphism is not part of the definition of x^n, so you cant state that without proof.

[Anonymous No.6465360]

because a/a=n^a/n^a=n^0=1

[Anonymous No.6465372]

You can see x^n as a product of x's indexed by a set of n elements.
When n = 0, it's a product indexed by an empty set, hence the neutral element of multiplication, 1

[Anonymous No.6465378]

>>6465300
because we make math, and we can say whatever the fuck we want, who is going to stop us?

[Anonymous No.6465397]

what about 0^0???

[Anonymous No.6465398]

>>6465397
sometimes it's undefined, other times (particularly in algebra) it's defined as 1 (as a^b is defined as the number of functions from {1,2,....,a} to {1,2,....,b] and the number of functions from {} to {} is 1)

[Anonymous No.6465492]

>>6465356
Ahh, don't you just love pretentious people who spew out incoherent terms to try to confuse people and are wrong about it?

[Anonymous No.6465505]

>>6465492
>what is a joke

[Anonymous No.6465553]

>>6465356
Strictly speaking you can 'define' an equivalent and pointlessly obtuse definition for n^x for integer x by the homomorphism between the groups, that equals n at x=1, 1/n at x = -1. The identity mapping to the identity is a consequence but I think the whole joke is that this group theory version is hardly 'quite simple'/needless obfuscation/basically pointless.

[Anonymous No.6466649]

Because powers are subtracted when you divide by a power of the same base. For example:
>x^5 divided by x^3 is x^2, (because 5-3=2)

Therefore, to get a power of zero, you would do:
>x^5 divided by x^5 is x^0, (as 5-5=0)

Any non-zero number divided by itself is 1, so for all values of x in x^0, it is equal to 1.

[Anonymous No.6466697]

>>6465334
>>6465360
/thread

[Anonymous No.6466717]

In n^(1/x) x tends to infinity, meaning the term (1/x) becomes closer to 0 in turn giving you n^0

n^(1/4) = 4th root of n
n^(1/16) = 16th root of n
n^(1/inf) = inf'th root of n

As you can imagine, the xth root of n for a very large of x will end up close to 1, regardless of how small or large n is. As x approaches infinity, the x'th root of n will become 1.

[Anonymous No.6466904]

>>6466717
>>6465333
>>6465334
>>6465306
>>6465326
>>6465360

Just no.
n^0=1 (commonly) because we defined it to be so.
There are reasons to define it as 1, and it is compatible with other operations, but it's not the answer to the question, just nice effects of the definition.

[Anonymous No.6466971]

>>6465378

>we make math

HAHAHAHAHAHAHAHA.

[Anonymous No.6467081]

>>6466904
>just nice effects of the definition.
You're an idiot.

[Anonymous No.6467103]

assuming: n^(x+y) = (n^x)*(n^y)
(like (5^2)*(5^3)=5^(2+3)=5^5)

so: n^x = n^(x+0) = (n^x)*(n^0)
therefore n^0 = 1

[Chaeris No.6467123]

Exemple:

2^3 = 8
2^2 = 4

2^2 = (2^3)/2

Therefor:

2^1 = (2^2)/2 = 4/2 = 2

And the last step:

2^0 = (2^1)/2 = 2/2 = 1


And yes, you can pick any number to get the same answer.

Bonus:

2^(-1) = (2^0)/2 = 1/2
2^(-2) = (2^(-1))/2 = 1/2/2 = 1/4
And you can go on forever.

This proof is IMO the most simple yet the most effective one to the famous n^0=1

[Anonymous No.6467166]

>>6467123
so when we do exponents it is base 1? top kek m8

[Anonymous No.6467338]

>>6465300
0^0=1 and 0^n=0

[Anonymous No.6467347]

>>6467081
I am mathematician. And yes, it is defined so that is has nice properties, like
>>6467123
It is not all that intuitive, since 0^n = 0 for all n!=0.

It makes the most sense to define it like this.


AND after all, in modern analysis it is often the case that at least 0^0 is left undefined.

So, before calling someone an idiot, use your brain or if not able, at least use google.

[Anonymous No.6467353]

>>6467347
It was meant to say " like, for example 0! = 1.

[Anonymous No.6467508]

2^5 = 32
2^4 = 16
2^3 = 8
2^2 = 4
2^1 = 2
2^0 = 1
2^-1 = 0.5
2^-2 = 0.25
2^-3 = 0.125

Do you see the pattern? It's divide by 2 everytime

[Anonymous No.6469415]

>>6465333
you're dumb. stop trying to look smart you fckin undergrad

[Anonymous No.6469460]

>>6465398
It's not 1 when looked at through complex numbers.

[Anonymous No.6469469]

Op, its because x^n-1 is equal to (x^n)/x
so x^n-1 for n=1 is x/x which is 1

[Anonymous No.6469482]

>>6465300
a=1*a for all numbers a, by definition of 1.
a^b = 1*a^b = 1(*a*a*a*...*a) with b (*a)'s.
n^0 = 1*n^0 = 1() with zero (*n)'s = 1.

0^0=1*0^0=1() with zero zeroes = 1.

[Anonymous No.6470533]

n^0 = (n^1/n^1) = 1.

[Anonymous No.6470538]

>>6470533
nice shitbump

[Anonymous No.6470540]

>>6470538
I don't swing that way gay boy.

[Anonymous No.6470561]

interdasting...now why is 0!=1?
1!=1 so 1=0 <=> you all lost.

[Anonymous No.6470587]

>>6466904
People like you who don't understand that the definition of n^0 to 1 is not the reason n^0=1 lack a key quality required to achieve anything meaningful in mathematics. Incidentally, if you cannot guess what that quality is, it's also because you're lacking it and it proves my point.

[Anonymous No.6470590]

>>6470561
"Every function is injective" -- Retarded Anon

[Anonymous No.6470970]

>>6470587
>you lack some unspecified intangible property and if you don't guess what it is you have proved my point
dis gotta be baitn

[Anonymous No.6471809]

>>6470561
because n!=(n+1)!/(n+1)