Math101 - Lecture 6 revision

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Maths 101 – Lecture 6 8 March 2000
f: X  Y x f Y 9 Z
g: Y  Z
Composition If f: x  y, g: Y  Z
And f(x) = y∈ Y for x∈ X
g(y) = ~∈Z
We define the function gxf by (g0f)(x) = g(f(x))
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Function x  z
gxf
X  Z
eg(i) Find (gxf) (x) if: -
h = (gxf)
f(x) = x2 + 1
g(x) = x – 1

The composition is (gxf)(x) = g(f(x)) = g(x2 + 1) = (x2 +1) – 1 = x2

Or
Let y = f(x) = x2 + 1
Then g(f(x)) = g(y) = y – 1 = (x2 + 1) – 1 = x2

What is (fo g)(x)2

(fo g)(x) = g(f(x)) = [g(x)]2 + 1 (remember f(x) = x2 + 1
= (x – 1)2 +1
= x2 – 2x + 2 ≠ (go f)

(2) find goh for h: IN  R and g: R  R
Given by h(h) = 1/n
g(x) = 1/(x2 + 1)

Solution (goh)(n) = g(h(n)) = 1/([h(n)]2 + 1
= 1/((1/n)2 +1)
= n2/(1 + n2)

Numbers
Natural or counting numbers: - 1,2,3, … use the symbol IN
Note: M + n ∈ IN for m, n ∈ IN
Nn ∈ IN for m, n ∈ IN} Closure under + and x

IN is ordered eg 1 < 2, or 3 > 2, etc
We need to extend the system IN, because we cannot always solve equations, m + x = n, within IN.

Since if m > n there is no solution with x ∈ N we extend IN to include 0 and “negatives” of the elements of IN. We have the integers Z

Integers Z = {…, -2, -1, 0, 1, 2, 3, … }
= {x: x ∈ IN or –x ∈ IN}2

Properties (above) still hold x m + n ∈ Z m, n Z
M, n Z, m n2 } Closure under +, x
Order,, + < 0

In addition we also have an additive identity element 0 Z
M + 0 = 0 + m = m for any m Z
(Still have multiplicative identity 1, m x 1 = 1 x m = m
We need to extend Z if we are to solve equation mx = n for m, n ∈ Z eg if m = 2 and n = 1 then we cannot have x ∈ Z.
We extend to the rationals
The rationals , Q { x:x = ρ/q, q ≠ o, p, q ∈ Z}


Q ∪ Z ∪ IN


Properties still hold: r + s ∈ Q for r, s ∈ Q
} Closure under +, x
rs ∈ Q for r, s ∈ Q
identities 0, 1
order, need to be careful.

We define a/b < c/d, for a/b, e/d ∈ Q (a, b, c d ∈ Z)

For b > 0, d > 0 by ad < bc

Note: we can always take b > 0 in a/b, since we can write a/b > -a/-b if b < 0
eg Is ½ > 1/3 ?
Solution: ½ > 1/3 means 1*3 > 2 * 1 which is true.