# [Notes](#fn1): Problem Class, 26 Oct 2020
> Group 11 -- Problem Class, MATH40001/40009 IUM
> Stergios M. Antonakoudis (stergios@imperial.ac.uk)
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Welcome to IUM Problem Class Group 11. \
Today's class: <u>**Problem Sheet 4 of Part II.**</u>
[^1]: <small> *These notes[^1] are typed up and created during and for the purposes of the discussion in the online session of Problem Class Group 11 for IUM on MS Teams, similarly to the use of a 'whiteboard' in an actual class.* </small>
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### $\sqrt{2}$ is not a rational number
### Proof (using geometry)
Arguing by contradiction, let's assume that $\sqrt{2}= \frac{B}{A} \iff B^2 = 2 \cdot A^2$, where $A,B$ are ==least== natural numbers ==with these property==.
==We assume that $B$ is the smallest natural number such that there exists a natural number $A$ with $B^2 = 2 \cdot A^2$.==
There is a square of side-length $B$, whose **area** is twice the area of a square of side-length $A$. We assume that both squares of integer side-lengths and $B$ was the least *integer square* with this property.
Consider two squares with side-lengths $B' = 2A -B$ and $A'= B-A$. Check that $(B')^2 = 2 \cdot (A')^2$. A contradiction!
(Picture)
*QED*
### Problem X' (extra)
Find a *similar* proof that $\sqrt{3}$ is not a rational number.
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## Problems
### ==Problem X (extra)==
Given $a,b \in \mathbb{R}$. Consider the sequence of numbers obtained as follows:
$a_1 = a,\;\; a_2=b,\;\;, a_n = |a_{n-1}| - a_{n-2}$ for $n \geq 3$. Find $a_{2020}$.
### ==Solution X (sketch)==
What happens if a=0 and b=1? -- ==0,1==, **1, 0**, -1, 1, 2, 1, -1, ==0, 1==
What happens if a=1 and b=0? -- **1,0**,...,1,0 a loop of *length 9*!
==Prove it always has a loop of length 9!==
> E.g. $a_{10} = a$ always!
*QED*
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