# CMPUT272 Lecture Notes - Lecture 19: Binary Logarithm

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November 13 2014
ETLC 2-001
f B is a function a!b f
Function f: A -> B is 1-1 a1,a2 a1 = a2
is onto ba C
gaf(a) = g(f(a))
Example:
A={1,2,3,4}
B={a,b,c}
C={a,b,c,d}
D={5,6,7,8,9}
f={(1,a),(2,a),(3,a),(4,b)}
g={(a,9),(b,8),(c,7),(d,5)}
gg = undefined
Just wrong
f={(1,c),(2,c),(3,c),(4,b)}
g={(a,9),(b,8),(c,7),(d,5)}
gg = undefined
Example:
a:
a: (x,y) -> x + y
r4: a: a: -> r4(a(x,y))
Theoroms: For functions f: A -> B and g: C -> D where B f is
also 1-1
Suppose f and g are 1-1 and ga1,a2 a1 = a2
Let a1, a2 f(a1) = g A, f(a1) = f(a2) f is also onto.
A, f(a1) = f(a2) f is also onto.
Suppose B = C and f and g are onto.
Suppose B = C and f and g are onto.
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