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Let's suppose A rhymes with B and B rhymes with C. Does A always rhyme with C?

mike3996
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    First, I believe "transitional" should be "transitive". Second, I believe the answer is "yes", if include only "true" rhyme, that is, omit "forced" rhyme, such as rhyming "n" and "m", which is done a lot. – Hexagon Tiling Apr 06 '12 at 07:56
  • @HexagonTiling: good one! What were I thinking.. – mike3996 Apr 06 '12 at 09:57
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    Well ... Lord rhymes with fraud in England, and fraud rhymes with nod in California. – Peter Shor Apr 06 '12 at 10:18
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    @Hexagon Yes, many rhymes are just "close", not "exact". It's not really like a mathematical "equals" operation, but more like "approximately equal". So just as you could say that 1.00 ~= 1.01 and 1.01 ~= 1.02, after a long enough chain you could end up with something quite different from what you started with. – Jay Apr 06 '12 at 15:04

3 Answers3

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Except in the technical case of equivocal words (where a single word has multiple pronunciations). If B was an equivocal word, it could rhyme with both A and C, yet A and C would not rhyme. For example:

A = taxes
B = axes
C = taxis

In this case, B is the plural of axe to rhyme with A, and the plural of axis to rhyme with C, but A and C do not rhyme.

(Some might argue that this shouldn't count, because B is really two words, not one. Fair enough; I did mention this counterexample was a technicality that breaks the transitive property of rhyming. I still felt it was worth a mention.)

EDIT. Since I've been called on my counterexample in the comments below, I offer one more example to mull over:

A = dollop \ˈdä-ləp\
B = scallop \ˈskä-ləp, ˈska-ləp\
C = gallop \ˈga-ləp\

Most dictionaries list two pronunciations for scallop; one rhymes with A (dollop), the other rhymes with C (gallop).

Now, B isn't technically two words, but one - albeit with two pronunciations. Still, A rhymes with B, and B rhymes with C, but A doesn't rhyme with C.

So, as the O.P. asked, "Does A always rhyme with C?"

J.R.
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    Thanks. I was skeptical about the transitiviness and thought there must be a clear counterexample lying there. Well, this is one even if a bit technical. – mike3996 Apr 06 '12 at 10:00
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    I don't feel this is a counterexample. As JR himself/herself has said, "axes" (more than one axe) and "axes" (more than one axis) are two different words. –  Apr 06 '12 at 10:27
  • @progo: I agree with David Wallace that this is not a counterexample. I agree with the answer that he gave. However, I can't refrain from telling an old joke that hedges one's bets: "If A = B, and B = C, then A = C, except where void or prohibited by law." – Hexagon Tiling Apr 06 '12 at 11:32
  • @DavidWallace: if that's your qualm, we can fix it (see edit). – J.R. Apr 06 '12 at 15:04
  • Wow, that's a good edit. I hadn't even thought of that particular case. I guess rhyming isn't transitive after all. I would hypothesise that in any one person's idiolect, rhyming is transitive; although I suspect that if I do, someone will come along and say "well, sometimes I pronounce neither with an EE sound, and sometimes I pronounce neither with an EYE sound, so it's not true in my idiolect". Maybe I should delete my original answer. –  Apr 07 '12 at 07:42
  • @DavidWallace: I wouldn't delete your original post; it's all part of an intelligent conversation. My initial thought was the same as yours: "Of course rhyming is transitive – what a silly question!" But then I gave it some extra thought... blame it on some really good math instructors I've had through the years. – J.R. Apr 07 '12 at 08:30
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Yes. "A rhymes with B" is a relationship of sameness; that is, a certain part of A is the same as the corresponding part of B. The mathematical term for this is an "equivalence relation", which is always reflexive (everything rhymes with itself), symmetric (if A rhymes with B, then B rhymes with A) and transitive.

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    Yes, something I was thinking of. But there are more than enough ambiquities and other examples that I'm not sure if the transitiviness holds trivially. – mike3996 Apr 06 '12 at 10:21
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    Ambiquities ... sort of like iniquities? – GEdgar Apr 06 '12 at 15:05
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If you're talking about perfect rhyme, the answer is no. The pair obtain and remain are a perfect rhyme, as are retain and remain, but obtain and retain do not form a perfect rhyme, since they both end with -tain, and the vowels before -tain are different. If you want rhymes to be transitive, you have to allow identical rhymes (see link above) and say that a word rhymes with itself.

Peter Shor
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