Some examples on proving/disproving a function is injective/surjective (CSCI 2824, Spring 2015)

This page contains some examples that should help you finish Assignment 6.

(See also Section 4.3 of the textbook)

Proving a function is injective

Recall that a function F:Ato B is injective/one-to-one if

forall, x,uin A, F(x)=F(u) implies x=u .

To prove that a function is injective, we start by: “fix any with ” Then (using algebraic manipulation etc) we show that .
To prove that a function is not injective, we demonstrate two explicit elements and show that .

Example 1: Disproving a function is injective (i.e., showing that a function is not injective)

Consider the function

$f:mathbb{R}^2tomathbb{R} : (x,y)mapsto sqrt{x^2 + y^2}$ .

(This function defines the Euclidean norm of points in $mathbb{R}^2$ .) Recall also that $mathbb{R}^2 triangleq mathbb{R}timesmathbb{R}$ .

Claim: is not injective.

Proof

Note that $(1,2),(2,1)inmathbb{R}^2$ are distinct and f(1,2)=sqrt{5}=f(2,1) . Hence is not injective. QED.

Example 2: Proving a function is injective

Consider the function

$g:mathbb{R}^2to mathbb{R}^2 : (x,y)mapsto (2x-y, -x+2y)$ .

Claim: is injective.

Proof

Fix any $(x,y), (u,v)in mathbb{R}^2$ satisfying .
By definition of , we have .
The equality of the two points in $mathbb{R}^2$ means that their coordinates are the same, i.e.,
Multiplying equation (2) by 2 and adding to equation (1), we get .
Then , or equivalently, .
On the other hand, multiplying equation (1) by 2 and adding to equation (2), we get , or equivalently, .
Therefore .
This proves that is injective. QED.

Proving a function is surjective

Recall that a function F:Ato B is surjectiveonto if

forall, yin B, exists, xin A left(, y= F(x), right) .

To prove that a function is surjective, we proceed as follows:
- Fix any .
- (Scrap work: look at the equation . Try to express in terms of .)
- Write something like this: “consider .” (this being the expression in terms of you find in the scrap work) Show that . Then show that .
To prove that a function is not surjective, simply argue that some element of cannot possibly be the output of the function .

Example 3: disproving a function is surjective (i.e., showing that a function is not surjective)

Consider the function

$h:mathbb{Z}tomathbb{Z}: xmapsto x^2$ .

Claim: is not surjective.

Proof

Consider .
Suppose on the contrary that there exists such that , i.e., .
Then being even implies that is even, i.e., for some integer .
Then , implying that , which is impossible because is an integer and the square of an integer must also be an integer.
Hence there does not exists such that .
This shows that is not surjective. QED.

Example 4: disproving a function is surjective (i.e., showing that a function is not surjective)

Consider the absolute value function

h:mathbb{R}tomathbb{R}: xmapsto |x| .

Claim: is not surjective.

Proof

Note that for any in the domain , must be nonnegative.
On the other hand, the codomain includes negative numbers.
Hence is not surjective.

Example 5: proving a function is surjective

Consider again the function

$g:mathbb{R}^2to mathbb{R}^2 : (x,y)mapsto (2x-y, -x+2y)$ .

Claim: is injective.

Scrap work

Fix any in the codomain $mathbb{R}^2$ .
We want to find a point in the domain satisfying .
Note that if and only if .
This is equivalent to and .
We are going to express in terms of .
Note that the first equation implies .
Substituting this into the second equation, we get .
Rearranging to get in terms of and , we get $x=frac23 u +frac13 v$ .
Now we work on . The second equation gives .
Substituting into the first equation we get .
Rearranging to get in terms of and , we get $y=frac13 u + frac23 v$ .
Hence the input we want is $(x,y)=(frac23 u+frac13 v, frac13u +frac23 v)$ .

Proof

Fix any in the codomain $mathbb{R}^2$ .
Consider $(x,y)=(frac23 u+frac13 v, frac13u +frac23 v)$ .
Note that lies in the domain $mathbb{R}^2$ and

$begin{array}{rcl} g(x,y)&=&(2x-y, -x+2y) &=& (2(frac23 u+frac13 v) -frac13u +frac23 v, -(frac23 u+frac13 v) + 2(frac13u +frac23 v) ) &=&(u,v). end{array}$

This shows that is surjective.

Finding the inverse

Once we show that a function is injective and surjective, it is easy to figure out the inverse of that function. The inverse is simply given by the relation you discovered between the output and the input when proving surjectiveness.

Only bijective functions have inverses!

A very rough guide for finding inverse

If we are given a bijective function f:Ato B , to figure out the inverse of we start by looking at the equation y=f(x) . Then we perform some manipulation to express in terms of .