this post was submitted on 11 Oct 2023
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submitted 1 year ago* (last edited 1 year ago) by hypertown@lemmy.world to c/memes@lemmy.ml
 
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[–] Ghyste@sh.itjust.works 0 points 1 year ago (1 children)
[–] Prunebutt@feddit.de 0 points 1 year ago

It's not less of a meme than most of the other posts in this community.

[–] ThatWeirdGuy1001@lemmy.world 0 points 1 year ago* (last edited 1 year ago) (3 children)

I know I'm bad at math but I don't understand how 2x0=0 but 2^0=1

How are they different answers when they're both essentially multiplying 2 by zero?

Someone with a bigger brain please explain this

Edit: I greatly appreciate all the explanations but all they've done is solidify the fact that I'll never be good at math 😭

[–] Globulart@lemmy.world 0 points 1 year ago* (last edited 1 year ago) (1 children)

This isn't strictly speaking a proof, but it did help me to accept it as it demonstrates the function that makes it 1.

2^3 = 2x2x2

2^2 = 2x2

(2^3)/(2^2) = (2x2x2)/(2x2) = 2

= 2^(3-2)

In general terms:

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

If a and b are the same number this is x^0 and obviously (x^a)/(x^a) is one because anything divided by itself is 1.

Hope that helps

[–] hemmes@lemmy.world 0 points 1 year ago (1 children)

Yes, of course, obviously...JFC, what??

[–] Flumsy@feddit.de 0 points 1 year ago

That was pretty complicated, here is a simpler answer I hsve come up with:

1=(2x2x2)/(2x2x2)=2³/2³=2³⁻³=2⁰

If that makes sense to you...

[–] jendrik@discuss.tchncs.de 0 points 1 year ago (1 children)

subtracting one from Exponent means halving (when the base is two):

2⁴ = 16 2³ = 8 2² = 4 2¹ = 2 2⁰ = 1

It's a simple continuation of the pattern and required for mathemarical rules to work.

[–] uberrice@feddit.de 0 points 1 year ago* (last edited 1 year ago)

This is confidently wrong.

3^0 is also 1. 2738394728^0 is also 1.

Edit: just saw that technically you're correct - sure.

IF base 2, Exponent reduction equals to halving - dividing by 2.

For x^y reducing y by one is equal to dividing by x, then we have the proof it always works.

[–] iamkindasomeone@feddit.de 0 points 1 year ago

Its not the same. And theres proof, why.

[–] abcd@feddit.de 0 points 1 year ago

Where are my programmer buddies? 🤘

[–] Ilovethebomb@lemmy.ml 0 points 1 year ago (1 children)
[–] affiliate@lemmy.world 0 points 1 year ago

for anyone curious, here's a "constructive" explanation of why a^0^ = 1. i'll also include a "constructive" explanation of why rational exponents are defined the way they are.

anyways, the equality a^0^ = 1 is a consequence of the relation

a^m+1^ = a^m^ • a.

to make things a bit simpler, let's say a=2. then we want to make sense of the formula

2^m+1^ = 2^m^ • 2

this makes a bit more sense when written out in words: it's saying that if we multiply 2 by itself m+1 times, that's the same as first multiplying 2 by itself m times, then multiplying that by 2. for example: 2^3^ = 2^2^ • 2, since these are just two different ways of writing 2 • 2 • 2.

setting 2^0^ is then what we have to do for the formula to make sense when m = 0. this is because the formula becomes

2^0+1^ = 2^0^ • 2^1^.

because 2^0+1^ = 2 and 2^1^ = 2, we can divide both sides by 2 and get 1 = 2^0^.

fractional exponents are admittedly more complicated, but here's a (more handwavey) explanation of them. they're basically a result of the formula

(a^m^)^n^ = a^m•n^

which is true when m and n are whole numbers. it's a bit more difficult to give a proper explanation as to why the above formula is true, but maybe an example would be more helpful anyways. if m=2 and n=3, it's basically saying

(a^2^)^3^ = (aa)^3^ = (aa) • (aa) • (aa) = a^2•3^.

it's worth noting that the general case (when m and n are any whole numbers) can be treated in the same way, it's just that the notation becomes clunkier and less transparent.

anyways, we want to define fractional exponents so that the formula

(a^r^)^s^ = a^r^ • a^s^

is true when r and s are fractional numbers. we can start out by defining the "simple" fractional exponents of the form a^1/n^, where n is a whole number. since n/n = 1, we're then forced to define a^1/n^ so that

a = a^1/n•n^ = (a^1/n^)^n^.

what does this mean? let's consider n = 2. then we have to define a^1/2^ so that (a^1/2^)^2^ = a. this means that a^1/2^ is the square root of a. similarly, this means that a^1/n^ is the n-th root of a.

how do we use this to define arbitrary fractional exponents? we again do it with the formula in mind! we can then just define

a^m/n^ = (a^1/n^)^m^.

the expression a^1/n^ makes sense because we've already defined it, and the expression (a^1/n^)^m^ makes sense because we've already defined what it means to take exponents by whole numbers. in words, this means that a^m/n^ is the n-th square root of a, multiplied by itself m times.

i think this kind of explanation can be helpful because they show why exponents are defined in certain ways: we're really just defining fractional exponents so that they behave the same way as whole number exponents. this makes it easier to remember the definitions, and it also makes it easier to work with them since you can in practice treat them in the "same way" you treat whole number exponents.