let f: R->R be defined as f(x)=3x-4. show that f is 1 - 1 and onto

all linear equations are one to one if they are on a slant

show that f(a) = f(b) means is satisfiable only when a=b

like f(a)=3x-4 f(b)=3x-4

yes, that is the basic premise of it.

cool and onto?

\[Let, x _{1} x_{2} \in R \] f(x1) =f(x2) 3(x1)-4=3(x2)-4 therefore x1=x2 so f is one-one

the fact that R is to R and all the elements from R are in R should suffice

ok. and with the sets of integers,, will be the same?

Z into R ?

or Z into and onto Z?

let y be any element of codomain R such that y=f(x) y=f(x) y=3x-4 3x=y+4 x=(y+4)/3 f[(y+4)/3]=3((y+4)/3)-4 =y therefore f is onto

well in the same question it is asking me,, now let g:Z->Z be defined by g(x)=3x4... is g 1 - 1 why and why not .. and id g onto .. why and why not

wow,, diyadiya... thats good

proofs aint me strong point

jajajaj,,, we r the same

and all my classes are proof :( im dying

so in the second one, should i do the same thing,,, they are integers :)

is \[g(x)=3x^4\]?

sorry, g(x)=3x-4

if you are mapping Z to Z g(x) is one-to-one ... do that was done above. but g(x) is not onto since g(x)=0 3x-4=0 3x=4 x=4/3 which is not an integer

ohhhh

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