# Bijective

• March 20th 2006, 04:49 PM
suedenation
Bijective
Hi, does anyone how to solve the following problems:

In each of the following cases, determine if the given function is bijective. If the function is bijective, find its inverse.

(a) The function f : R -> R defined by f(x) = 2x-3
(b) The function f : Z -> Z defined by f(x) = 2x-3

I don't really know what's the difference between (a) and (b).
I've done part (a), anything wrong?

Ans: (a)
Let yER, then if xER, then
y is the image of x under f <=> f(x)=y
<=> 2x-3=y
<=> x=(y+3)/2
so each yER is the image under f of a unique image of xER.
Therefore, f is bijective and has an inverse (y+3)/2

Thanks very much!!!! :D
• March 20th 2006, 06:22 PM
ThePerfectHacker
Quote:

Originally Posted by suedenation
Hi, does anyone how to solve the following problems:

In each of the following cases, determine if the given function is bijective. If the function is bijective, find its inverse.

(a) The function f : R -> R defined by f(x) = 2x-3
(b) The function f : Z -> Z defined by f(x) = 2x-3

I don't really know what's the difference between (a) and (b).
I've done part (a), anything wrong?

Ans: (a)
Let yER, then if xER, then
y is the image of x under f <=> f(x)=y
<=> 2x-3=y
<=> x=(y+3)/2
so each yER is the image under f of a unique image of xER.
Therefore, f is bijective and has an inverse (y+3)/2

Thanks very much!!!! :D

a)Prove it is bijective. First it is a function (TRUE). Next, it is one-to-one (injective) because if, $f(a)=f(b)$ then, $2a-3=2b-3$ thus, $a=b$ (TRUE). Next, we need to show it maps $\mathbb{R}$ onto $\mathbb{R}$. Which means $\forall,x\in\mathbb{R}\exists,y\in\mathbb{R}$ such as, $x=2y-3$. Which, is true because if you take $y=\frac{x+3}{2}$. Thus, $f:\mathbb{R}\to \mathbb{R}$ defined as $f(x)=2x-3$ is a bijective map. By, the property by bijective maps we replace the ordered pair $(x,y)$ for $(y,x)$ Thus, in $y=2x-3$ we have, $x=2y-3$ thus, $y=\frac{x+3}{2}$ is the inverse function.

b)Problem with this map is that it is not surjective. Meaning $\forall, x\in\mathbb{Z}, \exists, y\in \mathbb{Z}$ such as, $x=2y-3$. Now neccesarily true because $\frac{x+3}{2}$ can be a non-integer thus, $\frac{x+3}{2}\not \in \mathbb{Z}$

$\cal{Q}.\cal{E}.\cal{D}$