Math Memes

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Memes related to mathematics.

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NotAUser@lemmy.blahaj.zone

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

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

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[–] RagingNerdoholic@lemmy.ca 7 points 1 year ago* (last edited 1 year ago) (9 children)

I still can't wrap my head around why 0^0 = 1

If 0x0=0, and 0^0 is functionally identical to 0x0, why is it 1??

[–] Narrrz@kbin.social 4 points 1 year ago (3 children)

it makes graphs look nicer.

however, 0^0 isn't 0x0, that would be 0^2. 0^1 is 0x1, anything^0 is... well, it's 1. afaik there isn't am equivalent mathematical expression to n^0, it's multiplying a number by itself -1x, or something equally mind melting.

[–] 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.

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