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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