Kaleidoscope Mathematics part 1

Kaleidoscope Mathematics part 1, period 1a, year 1, University of Groningen, 2020-2021, the course is divided in 3 parts.
Kaleidoscope Mathematics deel 1, periode 1a, jaarr 1, Rijksuniversiteit Groningen, 2020-2021, het vak is in 3en gedeeld, document in het Engels.

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Kaleidoscope Mathematics
H.M. Goossens
September 2020
This course is divided in 3 parts. Modular arithmetic is the first one. Those are Lecture 1 till 5.
Lecture 1
1.1
Definitions:
Definition of integersaandb :We say a divides b (notation a|b) if integer c exists with b = ac.
Set of integers notation: Z
Some rules:
• a|b is an assertion, is not a number.
• b is a number.
• 2|6 is not a number, but assertion.
• 6|2 is false (because 2 = 6 · c) where c is not an integer, so 6|2 is false.)
• Notation false integer: 6 - 2
1.1.1
some examples:
When a|b then a|bc? For some integer a, b, c
Pf: by assumption integer d exists: b = a · d so bc = (ad)c = (dc)a
Because d and c are both integers: a|bc
1.2
Proof technique 1: Mathematical induction
If some integer a has a property and if it holds that n ≥ a having implies n + 1 also has it. And
with that, all n ≥ a have it.
1.2.1
Examples:
1. Example for every , ≥ 1 integer: 3|6n
Proof without induction:
6n = 3 · 2 · 6n−1
Here since n ≥ 1 we have 6n−1 is an integer so is, 2 · 6n−1
3|6n for all n ≥ 1 it holds for n = 1 since then 6 = 6 = 3 · 2 so it holds. Proof with induction:
Suppose for some n ≥ 1 −
→ 3|6n
This means 6n = 3 · a for some integer a
And hence 6n+1 = 6n · 6 = 3 · a · 6 = 3 · integer
Hence, proven by induction.
2. 43a n and x are both integers where n > 0 Suppose that x satisfies 3n|x + 1 . Show that this
implies that this implies that 3|x2 − x + 1 and also that 3n+1|x3 + 1
3n|x + 1 −
→ x + 1 = 3n · k
3α(3n−1 · k) = x + 1 −
→ 3|x + 1
x2 − x − 1 = (x − 1)(x − 1) − 3x. Because 3|x + 1 and 3|3x −
→ 3|x2 − x − 1
x+1=3n·k
3n|x + 1 −−−−−−→