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UNB/Year 4/Semester 2/CS2333/2024-01-17.md

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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