1. ## Injection and Surjection

Could you see if what i have done is right for the following questions:

Ive got to find whether they are injection surjection or bijection for any integer to any integer so z to z (sorry i cant use latex)

1. f(x) = x^3

bijection

2. g(x) = x-3

bijection

3. h(x) = 3x+1

bijection

4. i(x) = (x^2) -1

Surjection

2. First three are bijections, but the last one it's not, can you see why?

3. Originally Posted by Krizalid
First three are bijections, but the last one it's not, can you see why?
is it because i(n1) = i(n2)

one of them can be negative and one can be positive so n1 does not equal n2?

4. Meaning that for different points you got the same images, then yes.

5. Specfically, i(2)= 4+1= 5= i(-2).

6. Originally Posted by Krizalid
First three are bijections, but the last one it's not, can you see why?
No...1 and 3 are not surjective.

In 1, what maps to, say, 2? The image is only powers of 3, $\displaystyle \{ \ldots -9, -3, -1, 0, 1, 3, 9, 27, \ldots \}$.

In 3, what maps to, say, 2? Not every integer is of the form $\displaystyle 3x+1$ for $\displaystyle x \in \mathbb{Z}$...

7. Consider $\displaystyle x=\sqrt[3]2.$

8. Originally Posted by Swlabr
No...1 and 3 are not surjective.

In 1, what maps to, say, 2? The image is only powers of 3, $\displaystyle \{ \ldots -9, -3, -1, 0, 1, 3, 9, 27, \ldots \}$.

In 3, what maps to, say, 2? Not every integer is of the form $\displaystyle 3x+1$ for $\displaystyle x \in \mathbb{Z}$...

Editing a typo.

9. Originally Posted by Krizalid
Consider $\displaystyle x=\sqrt[3]2.$
But $\displaystyle \sqrt[3]2 \notin \mathbb{Z}$...

10. The original post is hard to read. But it does say that the functions are $\displaystyle \mathbb{Z}\mapsto\mathbb{Z}$.

12. Aaahh right i see.

My next question it says determine the inverse function wherever this exists.

How do we find out if an inverse function exists and then is it just 1 over the equation?

Aaahh right i see.

My next question it says determine the inverse function wherever this exists.

How do we find out if an inverse function exists and then is it just 1 over the equation?
No, the inverse function is NOT the reciprocal. I assume that by the time you are asked to find inverse functions, you have seen the definition and should know that. There is an unfortunate notation: we use $\displaystyle f^{-1}$ which, with numbers, implies the reciprocal (the multiplicative inverse) but the definition for functions is "$\displaystyle f^{-1}$ is the inverse function to f if and only if $\displaystyle f^{-1}(f(x))= x$ and $\displaystyle f(f^{-1}(x))= x$ for all x". That is, $\displaystyle f^{-1}$ "undoes" whatever f does to x and vice versa.

A standard way of finding an inverse for a function, f:
1) Write the equation y= f(x).
2) Swap x and y: x= f(y).
3) Solve for y.

For example, your first function if $\displaystyle f(x)= x^3$ so you would write that as $\displaystyle y= x^3$ and then swap x and y to get $\displaystyle x= y^3$. To solve for y, you would use the cube root, the "opposite" of cubing, to get $\displaystyle y= \sqrt[3]{x}$ or, equivalently, $\displaystyle y= x^{1/3}$.

However, these functions, we were told, were "from Z to Z". If that is still true, then this function does not have an inverse function because the cuberoot of an integer is not necessarily an integer.

14. So the inverse of y = x - 3
is
x = y - 3

y = x + 3

this is an inverse as all integers go to an integer

15. for 3

y = 3x + 1
x = 3y + 1
(x-1)/3 = y

y = (x-1)/3

not an inverse as it becomes a fraction not an integer when you put an integer in for x.

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