Question

Define g:R->Z by the rule
g(x)=9x+5
Is g(x) bijective function??
Prove or disprove

Answer 1

You can write down the defition of bijective function for g:R -> Z
1. f is injective function if for all x,y in R, if \[f(x_1)=f(x_2)\], then \[x_1=x_2\]
2. f is surjection if for all y in Z, there exists x in R such that \[y=f(x)\]

Answer 2

So kindly tell do you agree or disagree?? and what about g(x)=9x+5
What is confusing me is that for a function to be bijective, number of elements in R must be equal to number of elements in Z but R is an infinite set (correct me if i am wrong) and Z is not. So how can they be bijective? plus when I put values from domain i.e R in g(x) I do not get value i.e. output g(x) in the range of Z. Kindly please remove my confusion

Answer 3

No, Z is also infinite.

Answer 4

Ask yourself whether any integer can be produced from a real number given the function.

Answer 5

Bijective means that the function must be one-to-one. For every value of x there is exactly one value of y.

Answer 6

What you trying to prove here is that any value put into this function will remain an integer.

Answer 7

(1)So we are not concerned with the range of the output???
for example
put x=0.0005
the g(x)=9(0.0005)+5 is not the element of Z
So it is not a function

(2)but as all of you are saying function is bijective due to one to one correspondence between R and Z

which perspective should I follow (1) or (2)??

Answer 8

It doesn't matter that not all of the outputs are integers; it matters whether all integers can be produced.

Answer 9

O is that so!!!

Thank u very much blockcoder, sashley and ankh for ur help and time.