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A morpheme is the most smallest meaningful unit of language. A syllable is the smallest piece of pronunciation that has a vowel in it. Definitions are taken from this link.

I don't know why, but can we prove that mathematically or logically a morpheme would always have one or more syllables, but no syllable exist that contains more than one morpheme?

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    No, we can't prove such a thing because it isn't true. There are counterexamples everywhere, as the answers show. It is true, though, that monosyllabic words are usually special words in any language, if only because there is a limited number of possible syllables in any language. Even English has less than 4000 monosyllabic words.
    – jlawler
    Jul 29, 2016 at 3:08
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    There are zero/null morphemes so...
    – user0721090601
    Jul 31, 2016 at 11:23

5 Answers 5

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In English, one counterexample is the very common '-ed’ (often /d/) ending: ‘filled’ is 1 syllable, and the morphemes are ‘fill’ + ‘-ed’ (/d/).

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  • not that good example because it has confused spelling, but English has -s ending in both nouns and verbs.
    – Anixx
    Aug 4, 2016 at 9:14
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    Indeed, there are many such examples. And as others have mentioned, morphophonology is much more complex than this question would suggest. I’m not sure the spelling is particularly relevant, because the question is about syllables.
    – Jeremy Needle
    Aug 4, 2016 at 12:01
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It is perfectly possible to have three morphemes in one syllable. Consider the word sixths which is comprised of the morphemes /sɪks/, /θ/, and /s/.

So we can easily prove that many syllables exist that contain more than one morpheme.

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  • Thanks to Damkerng for sixths, which is a better word than the one I had before!
    – Araucaria - him
    Aug 3, 2016 at 15:48
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We can prove existentially that the shortest morpheme is a single consonant, Examples from Levantine Arabic: -ʃ "verbal negation"; -t "1sg perfective". In Gurage, single phonological features are morphemes.

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    How does this prove that the shortest morpheme is a single consonant? Your last sentence seems to disprove this (if we consider single phonological features to be "shorter" than single consonants).
    – brass tacks
    Jul 28, 2016 at 21:30
  • "Length" is the cardinality of the set of order indices, which is 1 in both cases. Single segments and features may differ in their complexity (number of dominating nodes, but they all have the same order index. So we don't actually consider single features to be shorter than conjunctions of (unordered) features.
    – user6726
    Jul 28, 2016 at 21:38
  • Oh, OK. Thanks for explaining. What would be the length of a "null morpheme" (assuming we accept that those exist)?
    – brass tacks
    Jul 28, 2016 at 21:50
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    It would depend on what they actually are. For example if they are just syntactic features and no phonological entry then phonological length is undefined. If they are a sequence of 1 or more empty P-feature matrices, then 1 or more. There's no known reason to posit the existence of something with length zero, so I would refrain.
    – user6726
    Jul 28, 2016 at 22:39
  • And how about tone morphemes, eg as in Hausa?
    – Gaston Ümlaut
    Jul 29, 2016 at 0:29
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In Spanish, the word "era" (was) can take no syllables, for example:

Adorarte para mi era obsesión

The part "mi era obsesión", when transcribed in IPA, would become /mi̯e.ɾao̯b.se.sjon/, wherein the "era" /e.ɾa/ doesn't increase the syllable-count.

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While many others have pointed out that there are many cases where multiple morphemes can exist in one syllable, it is also possible to have morphemes which in themselves do not constitute a syllable:

  • The copulative verbalizer in Aymara is indicated with elongation of the final vowel on the noun
  • The accusative case in Aymara is indicated with the deletion of the final vowel of the noun
  • Several American languages, like Mixtec, have a "floating nasal" morpheme, others indicate differences in e.g. person inflection with tonal contrast.

All examples above are morphemes, but not syllables.

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