~kb/open-notes

d6aa99277af15948d274d92e9b02b759c76457c4 — kb 10 months ago fb17222
CM1015 midterm WIP
1 files changed, 46 insertions(+), 22 deletions(-)

M notes/study/CM1015/CM1015-midterm.md
M notes/study/CM1015/CM1015-midterm.md => notes/study/CM1015/CM1015-midterm.md +46 -22
@@ 361,45 361,69 @@ $$

Suppose that $a$ and $b$ are integers where $a \equiv 11 \pmod{19}$ and $b \equiv 3 \pmod{19}$.

These congruences imply
$$
a = 19p + 11 \quad \text{and} \quad b = 19q + 3
$$

### i.

Given $c \equiv 13a \pmod{19}$ we may note
$$
\begin{align*}
c &= 19k + 13 \cdot 11\\
\\
&= 19k + 143
\\
\\
da &= p \cdot 19 + 11 \\

13a &= q \cdot 19 + r
c &\equiv 13a \pmod{19}\\
&\equiv 13 \cdot 11\\
&\equiv 143\\
&\equiv 7 \cdot 19 + 10\\
&\equiv 10
\end{align*}
$$


Given $c \equiv 13a \equiv 8b \equiv a - b \pmod{19}$ we may extrapolate
$$
c = p \cdot 19 + 13a
$$
### ii.

$$
c = q \cdot 19 + 8b
\begin{align*}
c &\equiv 8b \pmod{19}\\
&\equiv 8 \cdot 3\\
&\equiv 24\\
&\equiv 1 \cdot 19 + 5\\
&\equiv 5
\end{align*}
$$


### iii.

$$
p \equiv q \pmod n \implies p - q = k \cdot n
\begin{align*}
c &\equiv a - b \pmod{19}\\
&\equiv 11 - 3\\
&\equiv 8\\
&\equiv 0 \cdot 19 + 8\\
&\equiv 8
\end{align*}
$$

==TODO==

## Question 3 (b)

==TODO==
**Claim**  If $n$ is an odd positive integer, then $n^{2} \equiv 1 \pmod 8$.

_Proof._  This may be shown by direct proof.

By definition, if $n$ to be odd this implies $n = 2k + 1$ for some $k \in \mathbb Z$. 

From this we may state 
$$
\begin{align*}
n^{2} &= (2k + 1)^{2}\\
\\
&= 4k^{2} + 4k + 1\\
\\
&= 4k(k + 1) + 1
\end{align*}
$$
Given $2\ |\ k (k+1)\ \forall k$ it may be equivalently stated
$$
n^{2}= 8 \cdot m + 1, \quad m\in \mathbb Z
$$
As this form mirrors congruence to $1 \bmod 8$ it can be seen that for all odd positive integers the expressed relation holds.


## Question 3 (c)