# injectivity and surjectivity (onto and one-one)

• Feb 22nd 2009, 04:23 PM
scottie.mcdonald
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?
• Feb 22nd 2009, 04:32 PM
TheEmptySet
Quote:

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 $\displaystyle \frac{b+5}{4}$ the input is NOT an integer for all b.

suppose $\displaystyle 4n-5=2 \implies n=\frac{7}{4}$ 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.
• Feb 22nd 2009, 09:01 PM
benes
Quote:

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 ,$\displaystyle 4n-5\neq m$.

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 $\displaystyle 4n-5\neq m$.

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

A good problem is to find that A