52 lines
2.0 KiB
Markdown
52 lines
2.0 KiB
Markdown
Lecture Topic: Functions
|
|
|
|
For every input element $a \in A$ there is exactly one output element $b \in B$
|
|
|
|
Jargon:
|
|
- f maps a to b
|
|
- the image of a under f is b
|
|
|
|
functions are one-to-one (or injective):
|
|
$\forall x,y \in A, x \neq y \rightarrow f(x) \neq f(y)$
|
|
or in other terms
|
|
$\forall x,y \in A, x = y \rightarrow f(x) = f(y)$
|
|
|
|
Example:
|
|
$f(x) = x^2$ is not one to one
|
|
|
|
Proof by counter example:
|
|
$x = -1$ and $y = 1$ are two integers in the domain that $x=y$ but $f(-1)$ and $f(1)$ are both equal to 1, so it is not one to one
|
|
|
|
To prove a function is one to one, it is a little more tricky, and needs to be general, for example:
|
|
|
|
Let x and y be any arbitrary elements of the domain where f(x) = f(y)
|
|
So it follows that x and y are equal (x = y)
|
|
|
|
Example for a real function: $g(x) = 3x - 11$
|
|
Real Numbers
|
|
Let x and y be any real number such that $g(x) = g(y)$. We will show that $x = y$
|
|
Since we know that g(x) = g(y), it means that $3x - 11 = 3y - 11$
|
|
Add 11 to both sides: $\therefore 3x = 3y$
|
|
Divide both sides by 3: $\therefore x = y$
|
|
|
|
Important to note: The definition (domain and co-domain) can change if a function is one to one, for example the square function on all integers vs all positive integers
|
|
|
|
To prove that a function A -> B is onto:
|
|
- Let be represent and arbitrary element in the co-domain B
|
|
- We want to find an element a in the domain and show that f(a) = b
|
|
|
|
Example:
|
|
Real Numbers, $f(x) = 5x + 2$ is onto
|
|
|
|
Proof: Let b be any arbitrary real number (we want to find a real number a and show that f(a) = b)
|
|
|
|
A through process might follow that, I want a value a such that f(a) = b, I want $5a + 2 = b$, then $5a = b - 2$, then $a = \frac{b - 2}{5}$
|
|
|
|
So, let $a = \frac{b - 2}{5}$ , which is a real number. Then plug this back into the function
|
|
$f(a) = f(\frac{b - 2}{5}$) (Substitution)
|
|
$= 5\frac{b-2}{5}+2$ (Definition of f)
|
|
$= (b-2) +2)$ (Algebra)
|
|
$=b$ (Algebra)
|
|
|
|
Beginning of proving a function is not onto:
|
|
Find some specific co-domain element b and show that there is no domain element that could map to b |