Basic Modular Congruency Transitivity Proof

Question

Let it be said that I've never posted here, so hopefully I've formatted/asked okay.

The proof is as follows:

Let a, b, and c be integers and assume that n is an integer with n ≥ 2. Prove that if $a\equiv b\pmod n$ and $a\equiv c\pmod n$, then $b\equiv c\pmod n$. I understand that, by definition, this means that $n\mid (a -b) \;or\ (a-b)=nk$, where $k\in\mathbb{Z}$.

I found a Math Sorcerer video demonstrating its proof which I've screenshotted here, where, for the step where I was confused where to go next, he simply does: $a - b + b - c = nk_1 + nk_2$, where $k_1,k_2\in\mathbb{Z}$.

What's going on there? It seemed to me that he is simply assuming the "then" part of the statement as $(a-c)=nk$. The rest makes perfect sense to me, but it feels as though he's made a leap in logic there that I just don't follow.

See Modular arithemtic: we have that $(a-b)=k_1n$ and $(b-c)=k_2n$. Thus: $a-c=(a-b)+(b-c)=k_1n+k_2n=(k_1+k_2)n$. — Mauro ALLEGRANZA, Oct 12 '22 at 13:44
The details are: $a-c=(a+0)-c=(a+(b-b))-c=(a+(-b+b))-c=((a-b)+b)-c=(a-b)+(b-c)$ using obvious properties of numbers: associativity, commutativity, etc. — Mauro ALLEGRANZA, Oct 12 '22 at 13:57
@MauroALLEGRANZA Ah okay I see it now! Jeez I need a break apparently, thank you — Isaac Rosdail, Oct 12 '22 at 14:51

score 0 · Answer 1 · answered Oct 12 '22 at 13:44

0

There is no leap in logic. By assumption, you have that $a-b=nk_1$ and that $b-c=nk_2$ for appropriate integers $k_1, k_2$. Now note that $$ a-c = (a-b) + (b-c) $$ and this tells us that $$ a-c = nk_1+nk_2=n(k_1+k_2) $$ from which we deduce that $a \equiv c \bmod n$. Thus we deduce the condition for $a \equiv c \bmod n$, we never assumed it.

answered Oct 12 '22 at 13:44

Randall

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Wow I've done too much coding recently, my brain must be fried. I see what I was missing earlier, thank you! – Isaac Rosdail Oct 12 '22 at 14:49

Basic Modular Congruency Transitivity Proof

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