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Discrete Mathematics Functions

Functions in Group Theory with Examples

 

Functions

Let A and B be two non-empty sets. A function or a mapping f from A to be B is a rule which associates every element of A with a unique element of B and is denoted by f:ABf:A\to B .
In other words, a function f from A to B is a relation satisfying
i) Every element of A is related to some element of B.
ii) No element of A is related to two different elements of B.

 

Domain and Range of a Function

if f:ABf:A\to B is a function then A is called the domain and B is called the co-domain of f. If xAx\in A is associated with a unique element yBy\in B by the function f, then y is called the image of x under f and is denoted by y=f(x)y=f\left( x \right). Also, x is called the pre-image of y under f.

The range of f is the set of those elements of B which appear as the image of at least one element of A and is denoted by f(A)f\left( A \right). Thus f(A)={f(x)B:xA}f\left( A \right)=\left\{ f\left( x \right)\in B:x\in A \right\}. Clearly f(A)f\left( A \right) is a subset of B.

 Example 01

Let A = {1, 2, 3, 4} and Z be the set of integers. Define f:AZf:A\to Z by f(x)=2x+3.f\left( x \right)=2x+3. Show that f is a function from A to B. Also find the range of f.

Solution:

Now,

f(1)=5,f(2)=7,f(3)=9,f(4)=11

f={(1,5),(2,7),(3,9),(4,11)}

Since every element of A is associated with a unique element of B, f is a function.
Therefore Range of f={5,7,9,11}f=\left\{ 5,7,9,11 \right\}

 Example 02

Let N be the set of natural numbers. If f:NNf:N\to N is defined by f(x)=2x1f\left( x \right)=2x-1 show that f is a function and find the range of f.

Solution:

Now,

f(1)=1,f(2)=3,f(3)=5,.

f={(1,1),(2,3),(3,5),(4,7),.}

Clearly, f is a function.
Therefore, Range of f={1,3,5,7,....}f=\left\{ 1,3,5,7,.... \right\}

 Example 03

Let R be the set of real numbers. Defined f:RR  by  f(x)=x2f:R\to R\,\,by\,\,f\left( x \right)={{x}^{2}} for every xR.x\in R. Show that f is a function and find the range of f.

Solution:

Here f associates every real number to its square, which is certainly a real number. Hence f is a function. Range of R is the set of all non-negative real numbers.

 

One-One Function or Injection

A function f:ABf:A\to B is said to be one-one or injection if for all x1,x2A,f(x1)=f(x2)  implies  x1=x2.{{x}_{1}},\,{{x}_{2}}\in A,\,f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\,\,implies\,\,{{x}_{1}}={{x}_{2}}. The contrapositive of this implication is that of all x1,x2A,x1x2  implies  f(x1)f(x2).{{x}_{1}},\,{{x}_{2}}\in A,\,{{x}_{1}}\ne {{x}_{2}}\,\,implies\,\,f\left( {{x}_{1}} \right)\ne f\left( {{x}_{2}} \right)\,.

Thus a function f:ABf:A\to B is said to be one-one if different elements of A have different images in B.

 Example 04

Let R be the set of real numbers. Defined f:RR  by  i)f(x)=2x+3  ii)f(x)=x3f:R\to R\,\,by\,\,i)f\left( x \right)=2x+3\,\,ii)f\left( x \right)={{x}^{3}} for every xR.x\in R. Prove that f is one-one.

Solution:

i) Let f(x1)=f(x2)f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right) for some x1,x2R{{x}_{1}},\,{{x}_{2}}\in R

2x1+3=2x2+3

x1=x2

Thus for every x1,x2R,f(x1)=f(x2)  implies  x1=x2.{{x}_{1}},\,{{x}_{2}}\in R,\,f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\,\,implies\,\,{{x}_{1}}={{x}_{2}}. Therefore f is one-one.

ii) Let f(x1)=f(x2)f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right) for some x1,x2R{{x}_{1}},\,{{x}_{2}}\in R

x13=x23

x1=x2

Thus for every x1,x2R,f(x1)=f(x2)  implies  x1=x2.{{x}_{1}},\,{{x}_{2}}\in R,\,f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\,\,implies\,\,{{x}_{1}}={{x}_{2}}. Therefore f is one-one.

 Example 05

If f:RRf:R\to R is defined by f(x)=x2f\left( x \right)={{x}^{2}} for every xRx\in R , show that f is not one-one.

Solution:

Let

f(x1)=f(x2)

x12=x22

x1=±x2

Hence f is not one-one.
For example, f(2)=4  and  f(2)=4.f\left( -2 \right)=4\,\,and\,\,f\left( 2 \right)=4. The images of -2 and 2 are not different. Hence f is not one-one.

 

Onto Function or Surjection

A function f:ABf:A\to B is said to be onto or surjection if for every yBy\in B there exist at least one element xAx\in A such that f(x)=yf\left( x \right)=y i.e., every element of the co-domain B appears as the image of at least one element of the domain A.

If f is onto, then f(A)=Bf\left( A \right)=B.

 Example 06

Defined f:RR  by  i)f(x)=2x+3  ii)f(x)=x3f:R\to R\,\,by\,\,i)f\left( x \right)=2x+3\,\,ii)f\left( x \right)={{x}^{3}} for every xR.x\in R. Show that f is onto.

Solution:

i) Let yRy\in R then to find xRx\in R such that f(x)=y  i.e.,2x+3=yf\left( x \right)=y\,\,i.e.,\,2x+3=y
Solving for x we get, x=y32x=\frac{y-3}{2}
Since yR,x=y32Ry\in R,\,x=\frac{y-3}{2}\in R
Hence for every yR  exists  x=y32Ry\in R\,\,exists\,\,x=\frac{y-3}{2}\in Rsuch that f(y32)=yf\left( \frac{y-3}{2} \right)=y. Therefore f is onto.

ii) Let yRy\in R.
We shall show that there exists xRx\in Rsuch that f(x)=yf\left( x \right)=y.
That is x3=y.{{x}^{3}}=y.
Hence x=y13.x={{y}^{\frac{1}{3}}}. If yR,  then  y13R.y\in R,\,\,then\,\,{{y}^{\frac{1}{3}}}\in R.
Thus for every yRy\in R there exists y13R{{y}^{\frac{1}{3}}}\in R such that f(y13)=(y13)3=y.f\left( {{y}^{\frac{1}{3}}} \right)={{\left( {{y}^{\frac{1}{3}}} \right)}^{3}}=y.
Therefore f is onto.

 Example 07

If f:RRf:R\to R is defined by f(x)=x2f\left( x \right)={{x}^{2}} for every xRx\in R then prove that f is not onto.

Solution:

Since a negative number is the square of any real number, the negative numbers do not appear as the image of any element of R. For example, 9R-9\in R but there does not exist any xRx\in R such that f(x)=x2=9.f\left( x \right)={{x}^{2}}=-9. Hence f is not onto.

 

One-to-one function or Bijection

A function f:ABf:A\to B is said to be one-to-one or bijection if it is both one-one and onto.
For example, if f:RRf:R\to R is defined by i)f(x)=2x+3  ii)f(x)=x3i)f\left( x \right)=2x+3\,\,ii)f\left( x \right)={{x}^{3}} for every xRx\in R then f is one-to-one function.

 

Inverse Image of an element

Let f:ABf:A\to B be a function and yB.y\in B. Then the inverse image of y under f denoted by f1(y){{f}^{-1}}\left( y \right) is the set of those elements of A which have y as their image.

That is f1(y)={xA:f(x)=y}{{f}^{-1}}\left( y \right)=\left\{ x\in A:f\left( x \right)=y \right\}.

 Example 08

If f:RRf:R\to R\, is defined by f(x)=x23x+5f\left( x \right)={{x}^{2}}-3x+5 find (i)f1(3)  and  (ii)f1(15)\left( i \right)\,{{f}^{-1}}\left( 3 \right)\,\,and\,\,\left( ii \right)\,{{f}^{-1}}\left( 15 \right)

Solution:

i) Let

f1(3)=yf(y)=3

y23y+5=3

y23y+2=0

(y1)(y2)=0

y=1ory=2

Therefore, f1(3)={1,2}{{f}^{-1}}\left( 3 \right)\,=\left\{ 1,\,2 \right\}

ii) Let

f1(15)=yf(y)=15

y23y+5=15

y23y10=0

(y5)(y+2)=0

y=5ory=2

Hence, f1(15)={2,5}{{f}^{-1}}\left( 15 \right)=\left\{ -2,\,5 \right\}

 

Inverse Function

If a function f:ABf:A\to B is one-one and onto then the inverse of f denoted by f1:BA{{f}^{-1}}:B\to A is denoted by f1={(y,x):(x,y)f}{{f}^{-1}}=\left\{ \left( y,x \right):\left( x,y \right)\in f \right\}

Thus if f:ABf:A\to B is both one-one and onto then f1:BA{{f}^{-1}}:B\to A is obtained by reversing the ordered pairs of f.

Note: f1{{f}^{-1}} exists only when ff is both one-one and onto. Further f1{{f}^{-1}} is also one-one and onto.

 Example 09

Let Q be the set of the rational numbers. If f:QQf:Q\to Q is defined by f(x)=2x3f\left( x \right)=2x-3 for every xQx\in Q then find f1{{f}^{-1}} if it exists.

Solution:

i) Let

f(x1)=f(x1)

2x13=2x23

x1=x2

Hence f is one-one.

ii) Let yQy\in Q. Then to find xQ:f(x)=yx\in Q:f\left( x \right)=y
i.e., 2x3=y2x-3=y therefore x=y+32x=\frac{y+3}{2}
Whenever y is rational, x=y+32x=\frac{y+3}{2} is also a rational. Hence there exists y+32Q\frac{y+3}{2}\in Q such that f(y+32)=yf\left( \frac{y+3}{2} \right)=y
Hence f is also onto. Therefore f1:QQ{{f}^{-1}}:Q\to Q exists.

Let x=f1(y)f(x)=yx={{f}^{-1}}\left( y \right)\Rightarrow f\left( x \right)=y
i.e., 2x3=y2x-3=ytherefore x=y+32x=\frac{y+3}{2}
Define f1:QQ  by  f1(y)=y+32{{f}^{-1}}:Q\to Q\,\,by\,\,{{f}^{-1}}\left( y \right)=\frac{y+3}{2} for every yQy\in Q.
Replacing y by x we get f1(x)=x+32,  xQ{{f}^{-1}}\left( x \right)=\frac{x+3}{2},\,\,x\in Q. This is the required inverse function.

 

Composite Function or Product Function

If f:ABf:A\to Band g:BCg:B\to Care two functions then the composite function of f and g is denoted by gofgofis a function from A to C defined by (gof)(x)=g{f(x)}\left( gof \right)\left( x \right)=g\left\{ f\left( x \right) \right\} for every xAx\in A
Here (gof)(x)=g{f(x)}=g{y}=z\left( gof \right)\left( x \right)=g\left\{ f\left( x \right) \right\}=g\left\{ y \right\}=z

Note: goffoggof\ne fog. That is composition of mapping is not commutative.

composite function
 Example 10

Let R be the set of real numbers. Define f:RR  and  g:RR  by  f(x)=3x2  and  g(x)=x2+4f:R\to R\,\,and\,\,g:R\to R\,\,by\,\,f\left( x \right)=3x-2\,\,and\,\,g\left( x \right)={{x}^{2}}+4 find (i) gofgof (ii) fogfog

Solution:

(i)(gof)(x)=g{f(x)}=g{3x2}=(3x2)2+4=9x212x+8xR

(ii)(fog)(x)=f{g(x)}=f{x2+4}=3(x2+4)2=3x2+10xR

 Example 11

If f(x)=11x,x1f\left( x \right)=\frac{1}{1-x},\,x\ne 1 find f[f{f(x)}]f\left[ f\left\{ f\left( x \right) \right\} \right]

Solution:

Now,

f[f{f(x)}]

=f[f{11x}]=f[1111x]=f[1x1x1]=f[1xx]=f[x1x]

f[f{f(x)}]=11x1x=xxx+1=x

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