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Define a relation on $\mathbb N\times \mathbb N$ by $$ (a,b)\sim (c,d) \iff a+d=b+c. $$

Prove that if $(a, b)\sim(a', b')$ and $(c, d)\sim(c', d')$, then $(ac+bd,bc+ad)\sim(a'c'+b'd',b'c'+a'd')$

Here's what I've tried: Given that $(a,b)\sim(a′,b′)$ and that $(c,d)\sim(c′,d′)$, we see that $a+b′=b+a′$ and $c+d′=d+c′$. Thus, $a+b′=b+a\Rightarrow(a+b′)(c+d′)=(b+a′)(c+d′)\Rightarrow ac+ad′+b′c+b′d′=bc+bd′+a′c+a′d′\Rightarrow ac+ad′+b′(c+d′)=bc+bd′+a′(c+d′)\Rightarrow ac+ad′+b′(d+c′)=bc+bd′+a′(d+c′)\Rightarrow ac+ad′+b′d+b′c′=bc+bd′+a′d+a′c$

Jay
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1 Answers1

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Hint: You will probably find it significantly easier to prove this if you restate the (equivalence) relation as $$ (a,b) \sim (c,d) \iff a-b = c-d $$ Note that $$ ac + bd - (bc + ad) = (a-b)(c-d) $$

Ben Grossmann
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  • Thanks for the hint. I saw that this was possible, but I was worried about the case where $b > a$ or where $d > c$, since the relation is only defined on $\mathbb{N}\times\mathbb{N}$. Would this restatement of the equivalence relation cause any problems of that sort? – Jay Jan 16 '15 at 05:18
  • The relation, though defined on the natural numbers, can be defined using negative integers without any "problems". – Ben Grossmann Jan 16 '15 at 05:21
  • From an aesthetic point of view, it could be said that my answer requires "bringing the integers" into a problem that was posed perfectly well with just the natural numbers, so perhaps my answer is less than ideal (though logically correct). – Ben Grossmann Jan 16 '15 at 05:24
  • I see what you're saying, although I don't entirely understand why this is allowed. Is that because it's just a definition? – Jay Jan 16 '15 at 05:24
  • In fact, one might want to use this relation to define subtraction, in which case my proof would qualify as circular (within that line of reasoning, anyhow) – Ben Grossmann Jan 16 '15 at 05:27
  • My definition can be written as "the pairs of natural numbers $(a,b)$ and $(c,d)$ are related iff the integers $a-b$ and $c-d$ are equal". This is a perfectly acceptable definition that is demonstrably equivalent to the one given. – Ben Grossmann Jan 16 '15 at 05:31
  • @Ben Grossmann I am unfamiliar with the rules of arithmetic that would enable you to go from $ac + bd - (bc + ad)$ to (a-b)(c-d)$. Can you tell me what these rules are? – Edward.Lin Nov 19 '21 at 10:33
  • @Edward.Lin I don't know what kind of "rules of arithmetic" you're looking for, but one approach is factorization. $$ ac + bd - (bc + ad) = \ ac + bd - bc - ad = \ (ac - bc) + (bd - ad) =\ (a - b)c - (a-b) d =\ (a-b)(c-d). $$ – Ben Grossmann Nov 19 '21 at 17:11
  • Yes, but I don't know the rule underlying each of the steps in the equation (no high school maths). – Edward.Lin Nov 19 '21 at 19:05
  • @Edward.Lin I don't understand your question. Do you understand why each line is equal to the next? If not, then for which two lines do you not understand the equality? If yes, then what do you mean by the "rule underlying each step"? What do you mean by "no high school maths"? – Ben Grossmann Nov 19 '21 at 19:18
  • For the move from the first line to the second, it seems intuitive to me, but I couldn't explain why it is true. For the move from line 2 to line 3, it also seems intuitive, but I could not explain why it is true. For the move from line 3 to 4 and 3 to 4, I have no idea. – Edward.Lin Nov 19 '21 at 20:05
  • @Edward.Lin I see. The move from lines 1 to 2 simply the distributive property of addition. Note that $$ A + B - (C + D) = A + B + (-1) \cdot (C + D) = A + B + -C + -D = A + B - C - D. $$ The move from lines 2 to 3 uses a combination of the commutative and associative properties of addition. The moves between lines 3 and 4 and 4 and 5 are both applications of the distributive law, but "in reverse" (for instance, $ac - bc = (a-b)c$). – Ben Grossmann Nov 19 '21 at 20:09
  • What is the rule justifying $+−(+)=++(−1)⋅(+)$? Where can I learn these rules? – Edward.Lin Nov 19 '21 at 20:13
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    @Edward.Lin I would suspect that these rules are discussed in any middle/high school textbook on basic algebra, at least in the context of the real numbers. For more general settings, the statement $(-1) \cdot X = -X$ can be proved using the field axioms or even the ring axioms. – Ben Grossmann Nov 19 '21 at 20:25