Question

Is the function f : Z-->Z defined as f(x)=[x/3] one to one? I get that it's one to one because every x has a unique value, but I don't see how I can prove this Is the function f : Z-->Z defined as f(x)=[x/3] one to one? I get that it's one to one because every x has a unique value, but I don't see how I can prove this @Mathematics

Answer 1

Also what does f:Z-->Z mean that it's all real numbers?

Answer 2

z = all int numbers.

Answer 3

So is it still not one to one with the same reason

Answer 4

it is not 1 to 1, because, e.g when x=4, f(x)=4/3, which is not an integer number

Answer 5

x/3 = int number? yes, if x=1, 3, 6, ...
understand?

Answer 6

Oh I see, so only if it's an integer so the answer would be no because not all integers can have a unique integer value

Answer 7

;)

Answer 8

that is the greatest integer function right?

Answer 9

\[f(x)=[\frac{x}{3}]\] means greatest integer of
\[\frac{x}{3}\] so for example
\[f(4)=[\frac{4}{3}]=1\] and also
\[f(5)=[\frac{5}{3}]=1\]

Answer 10

Where does it say its the greatest integer function? because it has square brackets?

Answer 11

it is an function from integers to integers, but it is not a one to one function via the example i wrote above

Answer 12

square brackets means greatest integer function. also it says
\[f:\mathbb Z\rightarrow \mathbb Z\]

Answer 13

okay thanks! I was wondering what they were there for

Answer 14

so if it is not "greatest integer" it doesn't make any sense, because range would not be integers.

yw