Math Memes

1490 readers

5 users here now

Memes related to mathematics.

Rules:
1: Memes must be related to mathematics in some way.
2: No bigotry of any kind.

founded 1 year ago

MODERATORS

NotAUser@lemmy.blahaj.zone

151

Well, 0! =1, right? (lemmy.blahaj.zone)

submitted 1 year ago by urist@lemmy.blahaj.zone to c/mathmemes@lemmy.blahaj.zone

14 comments fedilink hide all child comments

you are viewing a single comment's thread
view the rest of the comments

[–] nekomusumeninaritai@lemmy.blahaj.zone 3 points 1 year ago* (last edited 1 year ago)

I'd imagine you want something defined recursively like multiplication

( 0x = 0 )
( xy = x(y-1)+ x ) ( y > 0 ).

So it needs to be

( x^0 = c ) (c is some constant)
( x^y = xx^{y-1} ) (( y > 0 ) (to see why, replace multiplication with exponentiation and addition with multiplication). So what could ( c ) be? Well, the recursive exponentiation definition we want refers to ( x^0 ) in ( x^1 ). ( x^1 ) must be ( x ) by the thing we wish to capture in the formalism (multiplication repeated a single time). So the proposed formalism has ( x = x^1 = xx^0 = xc ). So ( cx = x ) hence ( c = 1 ), the multiplicative identity. Anything else would leave exponentiation to a zeroth power undefined, require a special case for a zeroth power and make the base definition that of ( x^1 ), or violate the intuition that exponentiation is repeated multiplication.

On an unrelated note, it'd be nice if Lemmy had Mathjax. I just wrote all this on mobile with that assumption, and I'm not rewriting now that I know better.