Maths
101 – Lecture 9 Revision
Supermum x 
k
for x 
S
k
for x S
And
K -
x
for some x S
and > 0
x
for some x S
and > 0
Infinum k 
x,
for all x S
and k +> x, for some x S
and any > 0.
x,
for all x S
and k +> x, for some x S
and any > 0.
Eg 1
s1 = { x Q x ½
}
Q x ½
}
2 s2 = { x Q x2
< 2 }
Q x2
< 2 }
3
s3 = { ½, 2/3, ¾, ... n/(n + 1), ... }
4
s4 = { 1, ½, ¼, ... 2-n }
Which sets are bounded? What are the
supermum and infinmum?
Solutions
1 Bounded Supremum is ½ and is contained in S.
Infinmum is 0 and is contained in S.
2 Boundednote S2
= { x Q :
- √2 < x √2}
Q :
- √2 < x √2}
Supremum
is √2, note √2 is not in S2
Infinmum
is -√2, note √2 is not in S2
3 Bounded Supremum
is 1, note 1 is not in S3
Infinmum
is ½, note ½ is in S3
4 Bounded Supremum
is 1, note 1 is in S4
Infinmum
is 0, note 0 is not in
-//-
Complex numbers
Any equation x2 = a, a R
R
Cannot be solved for x R,
if a < 0
R,
if a < 0
Extend R so that x2 = -1 has a
solution
Yes can solve it in C (Complex numbers)
C is R with an element I = √-1 (and its
multiples etc)
Note R ⸦ C
Any complex number z C,
can be put into the form z = a +ib; a, b R
C,
can be put into the form z = a +ib; a, b R
a = Real part of z = a + ib, Re(z) = a
b = Imaginary part of z, a + ib, Im(z) = b
eg 2 – 3i C
C
Re(2-3i) = 2
Im(2-3i) = -3
Addition of elements C
Eg (2-3i) + (1+i) = 3 – 2i
Note: Re(z1 + z2) =
x1 + x2
Im (z1 + z2) = y1
+ y2
Multiplication
Eg (2 – 3i)(1 + i) = 2 + 2i – 3i – 3i2
=
2 – i – 3i2, but i2 =
-1
=
2 – i – 3 – 1
=
5 – i
General z1
= x1 + iy1
Z2
= x2 + iy2
z1z2 = (x1
+ iy1)( x2 + iy2) = x1x2 + ix1y2 + iy1x2
+ i2y1y2
= x1x2 – y1y2
+ i(x1y2 + x2y1)
(a)
(2
+ 3i)(10 – i) = 20 – 2i + 30i – 3i2
= 23
+ 28i
(b)
(1
+ i)2 = (1 + i)(1 =
i) (c) (√2 - i)2 = (√2)2 + (-i)2 -
2√2i
= 1 + i + i + i2 = 2 – 1 -
2√2i
= 2i =
1 - 2√2i
(d) i3 = i2i = -i
(e) i4 = (i2) = (-1)2
= i
(f) i5 = i ,
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