Thread: injectivity and surjectivity (onto and one-one)

I was going over my assignment to study for the upcoming midterm, and I thought I had a grasp on surjectivity and injectivity...I don't. Can someone tell me where I went wrong in proving these to 'jectives'.

define g:integers -> integers be defined by g(n)=4n-5. is it injective? surjective?

I wrote:

Injective:
Suppose g(n)=g(m)
4n-5=4m-5
4n=4m
n=m therefor it is injective

Surjective:
let f(a)=b
4a -5=b
a=(b+5)/4

sub back into the equation g(a)=4a-5 to get 4[(b+5)/4]-5
=b+5-5
=b, so therefor it is surjective

I think this is correct, but he wrote something along the lines of f(x)=2 has no solution in integers. This is true, so why does my way of doing surjectivity not work. What am I doing or assuming wrong?

Originally Posted by scottie.mcdonald

I was going over my assignment to study for the upcoming midterm, and I thought I had a grasp on surjectivity and injectivity...I don't. Can someone tell me where I went wrong in proving these to 'jectives'.

define g:integers -> integers be defined by g(n)=4n-5. is it injective? surjective?

I wrote:

Injective:
Suppose g(n)=g(m)
4n-5=4m-5
4n=4m
n=m therefor it is injective

Surjective:
let f(a)=b
4a -5=b
a=(b+5)/4

sub back into the equation g(a)=4a-5 to get 4[(b+5)/4]-5
=b+5-5
=b, so therefor it is surjective

I think this is correct, but he wrote something along the lines of f(x)=2 has no solution in integers. This is true, so why does my way of doing surjectivity not work. What am I doing or assuming wrong?

The Key is what set is your domain.

Your function is going from the integers to the integers

The statement the input is NOT an integer for all b.

Using your example

suppose but this is a rational number Not an integer.

So the function would be surjective over the rationals, but it is not over the integers.

I hope this helps.

Good luck.

Originally Posted by scottie.mcdonald

I was going over my assignment to study for the upcoming midterm, and I thought I had a grasp on surjectivity and injectivity...I don't. Can someone tell me where I went wrong in proving these to 'jectives'.

define g:integers -> integers be defined by g(n)=4n-5. is it injective? surjective?

I wrote:

Injective:
Suppose g(n)=g(m)
4n-5=4m-5
4n=4m
n=m therefor it is injective

Surjective:
let f(a)=b
4a -5=b
a=(b+5)/4

sub back into the equation g(a)=4a-5 to get 4[(b+5)/4]-5
=b+5-5
=b, so therefor it is surjective

I think this is correct, but he wrote something along the lines of f(x)=2 has no solution in integers. This is true, so why does my way of doing surjectivity not work. What am I doing or assuming wrong?

Proving surjectivity ( onto) for a function some times is very difficult if not impossible.

Remember the definition that the function g is surjective over integers is:

for all integers m we must find an integer n such that 4n-5=m

But for each definition you must consider its negation ,and the negation of the above definition is:

there exists an integer m such that for all integers n , .

If we put now m=1,we have: 4n-5=1 or n=6/4 .That result indicates that even if we try all the integers n we will not get 4n-5 equal to 1.

Hence the negation of the definition for surjectivity is satisfied,because:

there exists an integer m=1 such that for all n integers .

Thus g is not surjective over all integers ,but definitely is surjective over a
subset A of Z : g:

A good problem is to find that A