Math Memes

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Memes related to mathematics.

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

submitted 1 year ago by [email protected] to c/[email protected]

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[–] [email protected] 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.