- #1

member 587159

Let ##a,b,c \in \mathbb{N}##. If ##ac = bc##, then ##a = b##.

I thought, this was a pretty straightforward proof, but I think I might be doing something wrong.

Proof:

Let ##G := \{c \in \mathbb{N}|## if ##a,b \in \mathbb{N} ## and ##ac = bc##, then ##a = b\}##

Obviously, ##G \subset \mathbb{N}##. Since ##a.1 = b.1## implies that ##a = b, 1 \in G##.

Now, suppose ##g \in G##. Then ##ag = bg##. Thus, ##a.s(g) = b.s(g) \Rightarrow a(g+1) = b(g+1) \Rightarrow ag+a = bg + b \Rightarrow ag + a = ag + b \Rightarrow a = b## Thus, ##s(g) \in G## and we deduce that ##G = \mathbb{N}##. QED.

Also, I used some properties that I already proved:

##a.1 = a = 1.a##

##a + c = b + c \Rightarrow a = b##

The proof in my book however, defines ##G := \{a \in \mathbb{N}|## if ##b,c \in \mathbb{N} ## and ##ac = bc##, then ##a = b\}## and then uses some contradictions and other stuff to show that ##G = \mathbb{N}##. Am I overseeing something or is my proof correct as well? This page shows a similar proof to the one in my book: https://www.math.upenn.edu/~ted/203S10/References/Peano.pdf , theorem B.13.