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Mathematics 18 Online
OpenStudy (anonymous):

true or false? If f is a function, the f(s+t) = f(s)+f(t).

OpenStudy (aroub):

true

OpenStudy (anonymous):

depends on the function. thats not always true.

OpenStudy (anonymous):

Yeah, that's very often not true.

OpenStudy (anonymous):

counter example: \[f(x) = x^2\] f(x+y) doesnt equal f(x)+f(y) \[f(x+y) = (x+y)^2 = x^2+2xy+y^2\] \[f(x) = x^2, f(y) = y^2\]

OpenStudy (anonymous):

\(f(x) = x^2\) for a simple example. \[f(a+b) = (a+b)^2 = a^2 + 2ab + b^2 \ne a^2 + b^2\]

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

Hah, same example even..

OpenStudy (anonymous):

perfect! i got it!

OpenStudy (anonymous):

functions that do follow that rule are special. They are called homomorphisms. If they are invertible (bijective), then they are even better. those are called Isomorphisms. (to tie in what happened yesterday lol)

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