Help me

[Anonymous No.9221404]

I have never been taught base conversion, logarhythms, etc formally. I learned this TODAY.

Please help me understand how eucleudian division relates to this?
Please explain modulo and the base system?
Please explain how to convert from lower to higher bases, and vice versa?

>inb4 - 4 school
this is not for school. i genuinely want to know what all this shit means. I hunger for knowledge so please give me some.

pic related should be sign enough that i'm a brainlet who hasn't touched more than K-12 maths. Don't take me for a moron, i have somewhat of an idea of what is happening.

[Anonymous No.9221433]

>>9221404
bump for knowledge.

[Anonymous No.9222635]

>>9221404
Well you have to know how to use equivalence classes first. The modulo system is nothing les than: [math] X:=\{ [a] \, | \, a \in \mathbb{R} \}[/math] where the equivalence relation is defined by: [math]x \~ y : \iff x-y=k n[/math] for [math]k \in \mathbb{N}, n \in \mathbb{Z} [/math]. This is the way modn arithmetic is defined.

[Anonymous No.9222636]

>>9222635
and [math][a]:=\{x \in \mathbb{R} \, | \, x ~ a[/math]

[Anonymous No.9222640]

>>9222636[math] [a]:=\{x \in \mathbb{R} \, | \, x \sim a\}[/math]

[Anonymous No.9222650]

>>9221404
>logarhythms

[Anonymous No.9222666]

>>9222635
>Well you have to know how to use equivalence classes first. The modulo system is nothing les than: X:={[a]|a∈R}X:={[a]|a∈R} where the equivalence relation is defined by: x\~y:⟺x−y=knx\~y:⟺x−y=kn for k∈N,n∈Zk∈N,n∈Z. This is the way modn arithmetic is defined.
I don't know this notation too well.
>X:={[a]|a∈R}X
x is a where a is an element of real numbers (then an x on the end for some reason)?
>x\~y:⟺x−y=knx\~y:⟺x−y=kn for k∈N,n∈Zk∈N,n∈Z
please speak english, i'm new and brainlet.

[Anonymous No.9222704]

>>9222666
[math]x\sim y[/math] is called an equivalence relation if it satisfies these three properties:

i)[math]x\sim x[/math]
ii)[math]x \sim y \iff y \sim x[/math]
iii)[math]x \sim y[/math] and [math]y \sim z \iff x \sim z[/math].


Take for example the relation [math]x \sim y : \iff x-y \in \mathbb{Z}[/math]

i}[math]x-x=0 \in \mathbb{Z}[/math]
ii)[math] x-y=n \in \mathbb{Z}[/math] then [math]y-x=-(x-y)=-n \in \mathbb{Z}[/math]
iii)[math]x-y=n \in \mathbb{Z}[/math] and [math] y-z=k \in \mathbb{Z}[/math] then [math] x-z=x-y+(y-z)=n+k \in \mathbb{Z}[/math].
Now when you understand equivalence relations you can make a set containing all the elements that are equivalent to each other and call it an equivalence class noted as: [math] [a]:=\{ x \, | \, x \sim a \}[/math].

[Anonymous No.9222739]

>>9221404
Euclideand division:
Say you want to divide 45 by 8. You start counting multiples of 8 to see how far you can go before you go past 45. 1 times 8 = 8. 2 times 8 = 16, ..., 5 times 8 = 40, 6 times 8 = 48 WHOOPS that is already bigger.

So now you know that 45 = 5*8 + c where c is some number between 0 and 8. You can solve for c and get c = 5. 45 = 5*8 with remainder 5
Great jobs!

Mod:
Same as with division but now you hit your head against a wall enough times until you think that the remainder, c, is equal to the number itself. For example, the last paragraph showed that 45 = 5 mod 8 or in short that 45 = 5. If this does not make sense then hit your head again and try to see if now you get it.

Bases:
If you want to write m in base n then you need to write m as a linear combination of powers of n that uses coefficients that are only between 0 and n-1 inclusive.

For example. 15 = 1*10^1 + 5*10^0 in base 10
or 15 = 1*2^3 + 1*2^2 + 1*2^1 + 1*2^0 in base 2
or 15 = 1*3^2 + 2*3^1 in base 3

[Anonymous No.9223305]

>>9222739
Thanks anon.
btw, loved your explanation of mod

[Anonymous No.9223311]

>>9221404
I FUCKING HATE LOGS

[Anonymous No.9223384]

>>9221404
>pic related should be sign enough
It isn't. Not even people with math degrees get to touch the Laurent expansion of the stress energy tensor for a [math]N = 2[/math] supersymmetric affine conformal Lie algebra.

[Anonymous No.9223576]

>>9223384
Indeed. Mathematicians are generally not particularly interested in reading physics papers... unless they're written by people like Witten.

[Anonymous No.9223608]

>>9223311
[math]
\\ \displaystyle log_a \; x = \frac{log_b \; x}{log_b \; a}
\\
\\ \displaystyle lg\; 7 = \frac{ln \; 7}{ln \; 10}
[/math]