/sci/ - Science & Math

[Anonymous No.2618715]

Could someone please explain why 2^0 = 1 when calculating the decimal value of a binary number?

This shit never made sense to me and I can't find a proper explanation anywhere

[Anonymous No.2618726]

>>2618715
> can't find a proper explanation anywhere
> ask /sci/

[Anonymous No.2618724]

Because x^0 = 1 when x!=0

[Anonymous No.2618727]

In binary, zero equals one

[Anonymous No.2618732]

It follows logically. One law of exponents is a^b / a^c = a^(b-c)

2^8 / 2^8 = 2^(8-8) = 2^0 = 1

[Anonymous No.2618735]

anything^0 = 1

[Anonymous No.2618737]

Maybe because x^0 = 1 for all x != 0 is the only value that is compatible with desired exponentiation properties like
x^(a+b) = (x^a)*(x^b)

[OP No.2618759]

Still makes no sense to me, but thanks anyway guys

>mfw I try to math

[lol my !RIGjORdAn. No.2618762]

y= 2^x

[lol my !RIGjORdAn. No.2618768]

y=x^0

[Anonymous No.2618780]

any number to the zero power is 1.
the way i think about it is taking the number to some very small power. let's say you take it to the 1/20th power or the 1/100th power (taking the 20th root or 100th root). you will get an answer that converges on 1 as your fraction gets smaller and smaller.

[Anonymous No.2618785]

>>2618759
It is an abritrary rule that mathematicians decided on.
Like almost all of math.

[/sci/ has no gaems No.2618803]

I've never been completely satisfied with the way I was told but it's the most functional explanation I know of..

please brace yourself for a little repetition and rhetoric, just a little..

ok so when you have 2^2 you're essentially saying we have two factors of 2, two 2s multiplied together and when you have 2^3 you have three factors of 2, three 2s multiplied together or three 2s factored together..

alright, so, when you have 2^1 you have only one factor of 2. Your notation says you have only one 2 as a factor but factor at this point seems meaningless to people however it is still important semantically.

Now this is where the trick comes in, when you have 2^0 this does not mean you have zero 2s or zero factors, it means you have one less factor than 2^1 which is not zero because we are not adding and subtracting really; we're multipling and dividing. When you are going from 2^3 to 2^2 you're going (2^3)/2 = 2^2 = 4 and when you're going from 2^2 to 2^1 you're doing this operation: (2^2)/2 = 2^1 = 2. Therefore when you go from 2^1 to 2^0 you're doing this operation: (2^1)/2 = 2^0 = 1

further more when you go from 2^0 to 2^-1 you're likewise doing this operation: (2^0)/2 = 2^-1 = 1/2. Basically when you see the negative in the exponent you're dividing by whatever factor/term/expression however many times i.e. 2^-2 = ((1)/2)/2 = 1/4 & 2^-3 = (((1)/2)/2)/2 = 1/8.. in other, more simple words you just multiply those factors together like you normally do and then take the reciprocal.

Hope that helps!
recaptcha: judical from

[Anonymous No.2618811]

>>2618803

I sort of see the crazy ass logic train they have going with that, least it's easier to buy this shit now

Thankyou very much good sir

[Anonymous No.2618816]

>>2618785

Bullshit.

[/sci/ has no gaems No.2618827]

>>2618811
no problem, I love helping people out with explicitness.

[Anonymous No.2618834]

>>2618737
>>2618732
What the fuck is so hard for OP to understand about these explanations?

Additionally, take the limit of any f(x) = a^x + b graph as x approaches zero and it'll always equal 1 + b.

[Anonymous No.2618847]

>>2618834

Some of us cannot into math above minimum HS level

>Oh god what the fuck is this exclamation mark and shit, oh man I am so god damn confus

[Anonymous No.2619425]

>>2618732
To expand a bit on this post (because it doesn't explain why we choose to expand the definition of exponents to use that law):

So we invent this shit called "exponentiation." It's a symbolic representation of a*a*a*a*...n times...*a. Call this a^n. Cool.

*parties for a few years*

*is a math faggot philosopher*

*smokes pot*

Fag1: HEY DOODS WHAT IF WE WANT TO DO ^X, where X is not in a natural number?!

Fag2: WOAH YOU JUST BLEW MY MOIND

Fag1: OK, so what do we do for x^0?

Fag2: Well that can't mean x*0 times*x because that makes no sense! Let's try to derive it from a more formalized definition of powers. Hey! x^n/x^m = x^(n-m) so let's note x^0=x^(n-n)=x^n/x^n=1 so x^0=1

Fag1: AW HELLS YEAH MOTHERFUCKER

I make it a stupid convo, but that's basically how the development happened. Then you got into defininig it for all integers (the exact same way as for 0)

Then rationals using the laws that a^(1/n) = x where x^n=a (i.e., nth root defn)
and a^(m/n) = (a^m)^(1/n)

Then the rest of the reals using another definition, etc.

Most of math is about inventing shit that makes sense, then extending it over new groups/sets/fields to see what crazy shit you can do with it.

[Anonymous No.2619431]

>>2618785
>I suck at maths

[Anonymous No.2619459]

>>2619425
The easy defintion is just <span class="math">x^a = \exp \left(a \ln x\right) [/spoiler] for x positive.

[Anonymous No.2619483]

if 2^1 is 2, then dividing 2^1 by 2^1 will be 2^(1-1). But it's also dividing something by itself, which always produces 1.

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

[/sci/ has no gaems No.2621605]

I think this thread clearly emobodies /sci/'s communication problem. Understandably it's 4chan but you guys for the most part are too comfortable with implying things; this doesn't work well with explaining things to people you don't know and it's sad most of you won't give a shit but still call other people stupid (again, even when you don't know them).

>>2618735
this post is the epitomizes the communication problem. Nothing is more annoying when people pontificate what they've learned or have been taught. Not only does this fail at helping to explain "why", it's wrong..
as pointed out in >>2618737
..and arrogant on the verge of belligerent.

If nobody is really at fault though I think overall it's easy to simplify too aggressively. I know when I explain math to friends I forget what it's like to not know something and thus I fail repeatedly but those guys really don't know anything beyond division and multiplication so I have to reach deep-down/far-back (and keep everything short BUT not too short) if I ever want to succeed.

[Wittinyan !aN.EHCMpbY No.2621632]

The easiest to follow is:

2^3 = 2*2*2 = 8
2^2 = 2*2 = 4
2^1 = 2
2^0 = 2/2 = 1
2^-1 = 2/2*2 = 1/2
2^-2 = 2/2*2*2 = 1/4

[Wittinyan !aN.EHCMpbY No.2621657]

>>2621632

Actually, I should really have used brackets:

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

[/sci/ has no gaems No.2621666]

>>2621632
>>2621657
right but that could be too simple. Should be self-explanatory though..

..the only way to tell is see if other people who don't already know how to use exponents can understand and learn from that.

[Anonymous No.2621685]

You can get both 1 and 0, and you chose according to what kind of maths you are applying it to. Sorta undefined, but usually it's 1, at least for anything you are going to do.

[Anonymous No.2621790]

>>2621632
yeah, but that is just saying the same thing as >>2618735 really. he already knows 2^0=1, the question is why.

>>2618803
this one has at least some reasoning behind it.

[Anonymous No.2621922]

Is this still alive?! Fine, I'll expand on my post >>2618737

For positive integers, exponentiation can be viewed as repeated multiplication. For example, 2^5 = 2*2*2*2*2. It can easily be seen that 2^5 = (2*2*2) * (2*2) = 2^3 * 2^2. More generally: x^(a+b) = (x^a) * (x^b) for all integers a,b,x > 0.

Mathematicians then decided that it's useful to generalize the exponentiation operator to all positive real numbers, keeping the property that x^(a+b) = (x^a) * (x^b).

In particular, suppose that b = 0:
x^a = x^(a+0) = (x^a) * (x^0)

So either x^a = 0 for all x and a (this is obviously not a useful definition), or we must demand that x^0 = 1 for all positive x. You can use a similar idea to motivate exponentiating negative numbers.

[Anonymous No.2621953]

x^0 = x^(1-1)

x/x = 1 except when x=0

[Mathguy No.2622026]

OP, the explanation for that is this ( * stands for multiplies):
2^0 = 2^(1-1) = 2^1 * 2^-1
2^1 = 2
2^-1 = 1/2 = 0.5
2 * 0.5 = 1

There you go.