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

Question

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

amistre64 · Answer

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

amistre64 · Answer

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

anonymous · Answer

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

amistre64 · Answer

yes, that is the basic premise of it.

anonymous · Answer

cool
and onto?

diyadiya · Answer

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

amistre64 · Answer

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

anonymous · Answer

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

amistre64 · Answer

Z into R ?

amistre64 · Answer

or Z into and onto Z?

diyadiya · Answer

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

anonymous · Answer

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

anonymous · Answer

wow,, diyadiya... thats good

amistre64 · Answer

proofs aint me strong point

anonymous · Answer

jajajaj,,, we r the same

anonymous · Answer

and all my classes are proof :( im dying

anonymous · Answer

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

zarkon · Answer

is $$g(x)=3x^4$$?

anonymous · Answer

sorry, g(x)=3x-4

zarkon · Answer

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

anonymous · Answer

ohhhh