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For 44 claimed phonemes, we need 44*43/2 = 946 minimal pairs. If we can't find even one of them, then it is possible to claim that English has 43 phonemes and not 44 due to complementary distribution (with a complicated rule). Hence I'm interested if there is any work trying to count phonemes rigorously?

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    Generally, no, because Occam’s Razor still applies. You may not be able to provide any minimal pairs that prove beyond a shadow of a doubt that /a/ and /p/ are not complementarily distributed allophones of a single phoneme in English, but you’ll have a hell of a job convincing anyone that they are. The classic example is /h/ and /ŋ/, which are in (virtually) complementary distribution, but are still not seriously considered allophones by anyone. – Janus Bahs Jacquet May 3 at 20:52
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    Unfortunately, I can't undo my upvote for your comment @JanusBahsJacquet, but I don't think that Occam's Razor applies here. However, anti-intuitive it may be for /a/ and /p/ to be a single phoneme, it would suit Occam's Razor perfectly if they were. – Araucaria - he him May 4 at 0:59
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    Minimal pairs often form cliques, reducing the number of words to demonstrate them significantly, e.g., for English bail, sail, dale, fail, gale, hail, jail, mail, nail, pale, tail, whale, ... – jk - Reinstate Monica May 4 at 10:27
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In theory, any language could be analyzed as having only two phonemes, /0/ and /1/. Then we could say [p] is the realization of /00000/, and [t] is the realization of /00001/, and [k] is the realization of /00010/, and so on.

The problem is, this isn't an especially useful analysis. It doesn't reveal any new insights about the language, or make it easier to discuss sound changes, or anything like that.

Since phonemes are fundamentally a theoretical construct, not an externally-measurable one, what really matters is how useful they are. Do they explain the data in a concise way, and make it easier to analyze it? That's the true metric, more so than minimal pairs. Even though there's never going to be a minimal pair between [h] and [ŋ] in English (one only appears in the onset of a syllable, the other only in the coda), they act completely differently, and it doesn't really help anything to call them allophones.

For another illustrative example, look at different analyses of Mandarin vowels. Some linguists say that /y/ is phonemic in Mandarin, while others say it's merely an allophone of /ɥ/. But you'll never find a minimal pair, because linguists who prefer the first analysis say that every syllable needs a vowel in it, so you'll never see /ɥ/ alone without a vowel. (Linguists who prefer the second analysis say you can have syllables without phonemic vowels in them, and when you've got a /ɥ/ in a vowel-less syllable, you get [y].) Neither analysis is necessarily right or wrong, since both of them can explain all the data. So it comes down to elegance, utility, and personal preference.

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  • You can actually have a minimal pair between /h/ and /ŋ/, though it’s wordness is questionable: the derisive exclamation [hŋ̩˦˩] vs the exasperated exclamation [ŋ̩h˦˩]. Is that particularly useful, though? No, not really. That sort of context is exactly where you’ll also find minimal tonal pairs (‘hmm!’ vs ‘hmm?’), and nobody (sane) would claim that English is a tonal language. – Janus Bahs Jacquet May 3 at 22:56
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    @JanusBahsJacquet in that sort of context you also find things that aren't phonemes in a given language, such as clicks in "tsk"... – LjL May 4 at 3:02
  • This sounds similar to the question of whether physicists can produce a correct description of "reality". Physical theories are models, evaluated as to their usefulness in making predictions. Reality is a metaphysical issue, possibly unknowable. – Barmar May 4 at 15:44
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    If you really wanted to, I think you could analyze "song is" and "saw his" as a minimal pair between /ŋ/ and /h/. It's not always 100% clear where a syllable-break is, and the location of a syllable break often seems to be a consequence of the surrounding phonemes as much as a constraint on them. – ruakh May 5 at 0:14
  • @ruakh That example relies on an accent feature: the cot/caught merger. Without that merger, the vowel of "song" is one of English's short lax vowels that can't occur word-finally; and the vowel of "saw" is long, and as such cannot precede [ŋ]. (Modulo marginal unwordy English utterances.) – Rosie F May 8 at 9:50
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Even if you had a full set of minimal pairs, that actually would not rigorously establish the number of phonemes in English because it doesn’t tell you how the phonemes are segmented: you could make the number larger by treating phoneme sequences in the current analysis (like /sw/ /tw/ /dw/) as single phonemes, or smaller by treating single phonemes in the current analyses as sequences (e.g. [s] as /hz/).

In general, I don’t think it is common to seriously consider minimal pairs to be the primary practical or theoretical criterion for establishing a phoneme inventory. They are a useful way of demonstrating a phonemic contrast.

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  • this is not correct. For example, in English, sw tw dw cannot be considered phonemes, because, among other things, they do not occur word-finally, so they don't behave like regular phonemes, which normally can appear anywhere in a word (apart from the famous exception of h / ŋ̩). – Arnaud Fournet May 5 at 0:10
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    @ArnaudFournet: you just gave an example in your comment of an English phoneme that cannot occur word finally ... so clearly, it's not impossible for a phoneme to be absent in that position. Linguists disagree on whether to analyze "qu" in Latin as a sequence /kw/ or a labialized velar phoneme /kʷ/, even though "qu" cannot occur at the end of a Latin word (or syllable). It might not be a good idea to consider sw tw dw phonemes, but I think it's going a bit too far to say that this analysis can be definitely ruled out. – brass tacks May 5 at 1:08
  • As regards Latin, qu is clearly one phoneme and not a sequence of two phonemes k+u. This is shown by forms like lo"quitur with stress on o. If u in qu would count as a potential vowel, then it should be stressed loqu"itur. – Arnaud Fournet May 5 at 7:26
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    @ArnaudFournet Sure, kw acts different from other Cw clusters in Latin (it never splits across syllables), but that can still be explained away: maybe w acts different after velars because it's got a velar articulation itself. I personally prefer analyzing it as its own phoneme for a variety of reasons but it's possible to defend the other analysis too. – Draconis May 5 at 18:40
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    @ArnaudFournet Latin has a phoneme /w/, so the two-phoneme analysis would be that qu = /kw/, not /ku/; the stress of loquitur would be the same whether it's /lokwitur/ or /lokʷitur/. – TKR May 5 at 21:16
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As an addendum to the other answers, I'd like to observe that some languages lend themselves to minimal pairs, whereas others do not. This is a function of syllable structure, and probably to some extent the size of the phonetic/phonemic inventory as well. If all of your words have CV or CVC syllables, then it's going to be easy to find minimal pairs. If you allow CCVCCC, it's going to be harder. The fewer syllable types, the more languages are going to have to depend (so to speak) on words that are only minimally different.

I've experienced this personally: in an Iranian language I studied it was impossible or almost impossible to get minimal pairs for a contrast that was clearly phonemic, based on native speaker intuitions. And there were very few minimal pairs of any sort. On the other hand, in a Turkic language, I run into minimal pairs constantly, without even having to think about it.

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Phonemes, like morphemes, syllables, words, sentences are analytical tools. Like all tools, they are constructs of culture. A phoneme is no more real than the concept 'word.'

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You don't need to have minimal pairs that contrasts all phonemes with the others.
Phonemes are bundles of articulatory or phonetic features. If you have a pair that opposes p to t, and a pair that opposes t to d, and a pair for p vs b, then you don't need a pair for b vs d, because you already showed that labial is not dental, and voiceless is not voiced.

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    Doesn't this type of argument incorrectly predict that Arabic speakers would have a robust phonemic distinction between /b/ and /p/, since there is a distinction in Arabic between /t/ and /d/? But in fact, my understanding is that some Arabic speakers collapse /p/ and /b/ even in vocabulary of foreign origin, which seems to indicate that for such speakers the absence of /p/ in native vocabulary is not just an accidental gap, but corresponds to a gap in their phonemic inventory – brass tacks May 5 at 2:04
  • p does not exist in the first place (except in some dialects). – Arnaud Fournet May 5 at 7:27

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