Answer to Question #212054 in Discrete Mathematics for Sach

Question

Find out which of the following functions from R to R are (i) One-to-one, (ii) Onto, (iii) One-to-one corre￾spondence.

(a)

f: R—>R defined by f(x) = x

(b)

f: R—>R defined by f(x) = |x|

(c)

f: R—>R defined by f(x) = x + 1

(d)

f: R—>R defined by f(x) = x^2

(e)

f: R—>R defined by f(x) = x^3

(f)

f: R—>R defined by f(x) = x – x^2

(g)

f: R—>R defined by f(x) = Floor(x)

(h)

f: R—>R defined by f(x) = Ceiling(x)

(i)

f: R—>R defined by f(x) = – 3x+4

(j)

f: R—>R defined by f(x)= – 3x^2 +7

Answer 1

(a)

If $y_1=y_2,$ then $x_1=x_2=>$ One-to-one.

For every $y\in \R$ exists $x\in \R$ such that $x=y=>$ Onto

$f(x)=x$ is One-to-one correspondence (a bijection).

(b)

$$|-2|=2=|2|$$

If $y=-2,$ then there is no $x\in \R$ such that $|x|=-2.$

$f(x)=|x|$ is neither One-to-one nor Onto.

(c)

If $y_1=y_2,$ then $x_1+1=x_2+1=>x_1=x_2=>$ One-to-one.

For every $y\in \R$ exists $x\in \R$ such that $x=y-1=>$ Onto

$f(x)=x+1$ is One-to-one correspondence (a bijection).

(d)

$$(-2)^2=4=(2)^2$$

If $y=-4,$ then there is no $x\in \R$ such that $x^2=-4.$

$f(x)=x^2$ is neither One-to-one nor Onto.

(e)

If $y_1=y_2,$ then $x_1^3=x_2^3=>(x_1-x_2)(x_1^2+x_1x_2+x_2^2)=0$

$=>x_1=x_2=>$ One-to-one.

For every $y\in \R$ exists $x\in \R$ such that $x=\sqrt[3]{y}=>$ Onto

$f(x)=x^3$ is One-to-one correspondence (a bijection).

(f)

$$0-(0)^2=0=1-(1)^2$$

If $y=1,$ then there is no $x\in \R$ such that $x-x^2=1.$

$f(x)=x-x^2$ is neither One-to-one nor Onto.

(g)

$$Floor(0.2)=0=Floor(0.5)$$

If $y=0.5,$ then there is no $x\in \R$ such that $Floor(x)=0.5.$

$f(x)=Floor(x)$ is neither One-to-one nor Onto.

(h)

$$Ceiling(0.2)=1=Ceiling(0.5)$$

If $y=0.5,$ then there is no $x\in \R$ such that $Ceiling(x)=0.5.$

$f(x)=Ceiling(x)$ is neither One-to-one nor Onto.

(i)

If $y_1=y_2,$ then $-3x_1+4=-3x_2+4=>-3x_1=-3x_2$

$=>x_1=x_2=>$ One-to-one.

For every $y\in \R$ exists $x\in \R$ such that $x=\dfrac{-y+4}{3}=>$ Onto

$f(x)=-3x+4$ is One-to-one correspondence (a bijection).

(j)

$$-3(-1)^2+7=4=-3(1)^2+7$$

If $y=8,$ then there is no $x\in \R$ such that $x^2=8.$

$f(x)=-3x^2+7$ is neither One-to-one nor Onto.