Does a difference of tense count as a difference of meaning in a minimal pair?
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Does a difference in tense count as a difference in meaning in a minimal pair?
Here's a made up example to illustrate my question:
If we know that:
- [wuga] means "read"
- [wugi] means "reading"
Can we say that [wuga] and [wugi] are a minimal pair and that [a] and [i] contrast?
tense minimal-pairs
asked Mar 10 at 19:43
dawnchandlerdawnchandler
363
add a comment |
Does a difference in tense count as a difference in meaning in a minimal pair?
Here's a made up example to illustrate my question:
If we know that:
- [wuga] means "read"
- [wugi] means "reading"
Can we say that [wuga] and [wugi] are a minimal pair and that [a] and [i] contrast?
tense minimal-pairs
asked Mar 10 at 19:43
dawnchandlerdawnchandler
363
add a comment |
Does a difference in tense count as a difference in meaning in a minimal pair?
Here's a made up example to illustrate my question:
If we know that:
- [wuga] means "read"
- [wugi] means "reading"
Can we say that [wuga] and [wugi] are a minimal pair and that [a] and [i] contrast?
tense minimal-pairs
asked Mar 10 at 19:43
dawnchandlerdawnchandler
363
Does a difference in tense count as a difference in meaning in a minimal pair?
Here's a made up example to illustrate my question:
If we know that:
- [wuga] means "read"
- [wugi] means "reading"
Can we say that [wuga] and [wugi] are a minimal pair and that [a] and [i] contrast?
tense minimal-pairs
tense minimal-pairs
asked Mar 10 at 19:43
dawnchandlerdawnchandler
363
asked Mar 10 at 19:43
dawnchandlerdawnchandler
363
asked Mar 10 at 19:43
dawnchandlerdawnchandler
363
asked Mar 10 at 19:43
dawnchandlerdawnchandler
363
asked Mar 10 at 19:43
dawnchandlerdawnchandler
363
363
add a comment |
add a comment |
2 Answers
2
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oldest
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Yes. A minimal pair is meant to differ in one phoneme, to demonstrate that a speaker of the language can distinguish between the two words, and therefore that the contrast is phonemic. Since the difference between the words is on the level of phonology, it doesn't matter whether the difference in meaning is grammatical or lexical.
answered Mar 10 at 20:07
b ab a
1,7171326
add a comment |
Well, the issue is not as clear as the preceding answer states.
One issue here is morpheme-boundary.
It's not good policy to look for minimal pairs that involve morpheme boundaries. For example, drink drank drunk is ok because there is no boundary involved here, but there is the well-known case of Scottish English where vowel length normally does not exist, but some long vowels actually exist in words like Preterite laid < lay+ed. In other words, your pair wugi / wuga is not a fully satisfactory minimal pair. It might involve interferences caused by morpheme boundaries. So I would recommend to look for another minimal pair that does not involve that kind of issues.
answered Mar 10 at 21:30
Arnaud FournetArnaud Fournet
90827
2
Why would differences in morphological structure be relevant to deciding if two words form a minimal pair? The minimal pair test only refers to the phonetic segments in the words.
– user6726
Mar 10 at 23:38
2
I'm not sure I agree with this argument, but, for example, in English consonant length is generally not considered phonemic, yet if you take a word boundary, or even a morpheme boundary, you can have the same consonant twice, which results, for instance, in unaimed [ʌn'eɪmd] vs unnamed [ʌn'neɪmd], with a (potential, at least) geminate. Does this mean the analysis positing that consonant length is not phonemic in English is wrong, or does it mean that minimal pairs shouldn't be constructed across morpheme boundaries?
– LjL
Mar 11 at 0:22
1
@user6726 Rosa's vs. roses. This famous pair is not truly a minimal pair (Flemming & Johnson 2007). If English orthography marked the boundary between the stem and the plural suffix nobody would have even considered that possibility. Morphological structure is absolutely relevant because it shapes the definition of a word.
– Nardog
Mar 11 at 5:17
1
To follow Nardog, the issue is that defining what a "word" is not as clear as one may hope. So the best minimal pairs are those involving unanalyzable morphemes.
– Arnaud Fournet
Mar 11 at 10:08
@Nardog I've read the paper but it doesn't seem to say that Rosa's vs roses is not a minimal pair; I even find hints they consider it one, like "[...] a sequence of 25 minimal pairs (roses–Rosa’s, [...])". Anyway, while I think I see why you say they're not, and I've argued for that too earlier... if hypothetically Rosa's and roses were the only examples of [ɨ] vs [ə], would it be fair to say that /ɨ/ and /ə/ aren't separate phonemes in English because there is no minimal pair? I'd still say the utterances [ɹoʊzɨz] vs [ɹoʊzəz] show a meaningful distinction, indicating phonemicity.
– LjL
Mar 12 at 21:01
|
show 5 more comments
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2 Answers
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2 Answers
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Yes. A minimal pair is meant to differ in one phoneme, to demonstrate that a speaker of the language can distinguish between the two words, and therefore that the contrast is phonemic. Since the difference between the words is on the level of phonology, it doesn't matter whether the difference in meaning is grammatical or lexical.
answered Mar 10 at 20:07
b ab a
1,7171326
add a comment |
Yes. A minimal pair is meant to differ in one phoneme, to demonstrate that a speaker of the language can distinguish between the two words, and therefore that the contrast is phonemic. Since the difference between the words is on the level of phonology, it doesn't matter whether the difference in meaning is grammatical or lexical.
answered Mar 10 at 20:07
b ab a
1,7171326
add a comment |
Yes. A minimal pair is meant to differ in one phoneme, to demonstrate that a speaker of the language can distinguish between the two words, and therefore that the contrast is phonemic. Since the difference between the words is on the level of phonology, it doesn't matter whether the difference in meaning is grammatical or lexical.
answered Mar 10 at 20:07
b ab a
1,7171326
Yes. A minimal pair is meant to differ in one phoneme, to demonstrate that a speaker of the language can distinguish between the two words, and therefore that the contrast is phonemic. Since the difference between the words is on the level of phonology, it doesn't matter whether the difference in meaning is grammatical or lexical.
answered Mar 10 at 20:07
b ab a
1,7171326
answered Mar 10 at 20:07
b ab a
1,7171326
answered Mar 10 at 20:07
b ab a
1,7171326
answered Mar 10 at 20:07
b ab a
1,7171326
1,7171326
add a comment |
add a comment |
Well, the issue is not as clear as the preceding answer states.
One issue here is morpheme-boundary.
It's not good policy to look for minimal pairs that involve morpheme boundaries. For example, drink drank drunk is ok because there is no boundary involved here, but there is the well-known case of Scottish English where vowel length normally does not exist, but some long vowels actually exist in words like Preterite laid < lay+ed. In other words, your pair wugi / wuga is not a fully satisfactory minimal pair. It might involve interferences caused by morpheme boundaries. So I would recommend to look for another minimal pair that does not involve that kind of issues.
answered Mar 10 at 21:30
Arnaud FournetArnaud Fournet
90827
2
Why would differences in morphological structure be relevant to deciding if two words form a minimal pair? The minimal pair test only refers to the phonetic segments in the words.
– user6726
Mar 10 at 23:38
2
I'm not sure I agree with this argument, but, for example, in English consonant length is generally not considered phonemic, yet if you take a word boundary, or even a morpheme boundary, you can have the same consonant twice, which results, for instance, in unaimed [ʌn'eɪmd] vs unnamed [ʌn'neɪmd], with a (potential, at least) geminate. Does this mean the analysis positing that consonant length is not phonemic in English is wrong, or does it mean that minimal pairs shouldn't be constructed across morpheme boundaries?
– LjL
Mar 11 at 0:22
1
@user6726 Rosa's vs. roses. This famous pair is not truly a minimal pair (Flemming & Johnson 2007). If English orthography marked the boundary between the stem and the plural suffix nobody would have even considered that possibility. Morphological structure is absolutely relevant because it shapes the definition of a word.
– Nardog
Mar 11 at 5:17
1
To follow Nardog, the issue is that defining what a "word" is not as clear as one may hope. So the best minimal pairs are those involving unanalyzable morphemes.
– Arnaud Fournet
Mar 11 at 10:08
@Nardog I've read the paper but it doesn't seem to say that Rosa's vs roses is not a minimal pair; I even find hints they consider it one, like "[...] a sequence of 25 minimal pairs (roses–Rosa’s, [...])". Anyway, while I think I see why you say they're not, and I've argued for that too earlier... if hypothetically Rosa's and roses were the only examples of [ɨ] vs [ə], would it be fair to say that /ɨ/ and /ə/ aren't separate phonemes in English because there is no minimal pair? I'd still say the utterances [ɹoʊzɨz] vs [ɹoʊzəz] show a meaningful distinction, indicating phonemicity.
– LjL
Mar 12 at 21:01
|
show 5 more comments
Well, the issue is not as clear as the preceding answer states.
One issue here is morpheme-boundary.
It's not good policy to look for minimal pairs that involve morpheme boundaries. For example, drink drank drunk is ok because there is no boundary involved here, but there is the well-known case of Scottish English where vowel length normally does not exist, but some long vowels actually exist in words like Preterite laid < lay+ed. In other words, your pair wugi / wuga is not a fully satisfactory minimal pair. It might involve interferences caused by morpheme boundaries. So I would recommend to look for another minimal pair that does not involve that kind of issues.
answered Mar 10 at 21:30
Arnaud FournetArnaud Fournet
90827
2
Why would differences in morphological structure be relevant to deciding if two words form a minimal pair? The minimal pair test only refers to the phonetic segments in the words.
– user6726
Mar 10 at 23:38
2
I'm not sure I agree with this argument, but, for example, in English consonant length is generally not considered phonemic, yet if you take a word boundary, or even a morpheme boundary, you can have the same consonant twice, which results, for instance, in unaimed [ʌn'eɪmd] vs unnamed [ʌn'neɪmd], with a (potential, at least) geminate. Does this mean the analysis positing that consonant length is not phonemic in English is wrong, or does it mean that minimal pairs shouldn't be constructed across morpheme boundaries?
– LjL
Mar 11 at 0:22
1
@user6726 Rosa's vs. roses. This famous pair is not truly a minimal pair (Flemming & Johnson 2007). If English orthography marked the boundary between the stem and the plural suffix nobody would have even considered that possibility. Morphological structure is absolutely relevant because it shapes the definition of a word.
– Nardog
Mar 11 at 5:17
1
To follow Nardog, the issue is that defining what a "word" is not as clear as one may hope. So the best minimal pairs are those involving unanalyzable morphemes.
– Arnaud Fournet
Mar 11 at 10:08
@Nardog I've read the paper but it doesn't seem to say that Rosa's vs roses is not a minimal pair; I even find hints they consider it one, like "[...] a sequence of 25 minimal pairs (roses–Rosa’s, [...])". Anyway, while I think I see why you say they're not, and I've argued for that too earlier... if hypothetically Rosa's and roses were the only examples of [ɨ] vs [ə], would it be fair to say that /ɨ/ and /ə/ aren't separate phonemes in English because there is no minimal pair? I'd still say the utterances [ɹoʊzɨz] vs [ɹoʊzəz] show a meaningful distinction, indicating phonemicity.
– LjL
Mar 12 at 21:01
|
show 5 more comments
Well, the issue is not as clear as the preceding answer states.
One issue here is morpheme-boundary.
It's not good policy to look for minimal pairs that involve morpheme boundaries. For example, drink drank drunk is ok because there is no boundary involved here, but there is the well-known case of Scottish English where vowel length normally does not exist, but some long vowels actually exist in words like Preterite laid < lay+ed. In other words, your pair wugi / wuga is not a fully satisfactory minimal pair. It might involve interferences caused by morpheme boundaries. So I would recommend to look for another minimal pair that does not involve that kind of issues.
answered Mar 10 at 21:30
Arnaud FournetArnaud Fournet
90827
Well, the issue is not as clear as the preceding answer states.
One issue here is morpheme-boundary.
It's not good policy to look for minimal pairs that involve morpheme boundaries. For example, drink drank drunk is ok because there is no boundary involved here, but there is the well-known case of Scottish English where vowel length normally does not exist, but some long vowels actually exist in words like Preterite laid < lay+ed. In other words, your pair wugi / wuga is not a fully satisfactory minimal pair. It might involve interferences caused by morpheme boundaries. So I would recommend to look for another minimal pair that does not involve that kind of issues.
answered Mar 10 at 21:30
Arnaud FournetArnaud Fournet
90827
answered Mar 10 at 21:30
Arnaud FournetArnaud Fournet
90827
answered Mar 10 at 21:30
Arnaud FournetArnaud Fournet
90827
answered Mar 10 at 21:30
Arnaud FournetArnaud Fournet
90827
90827
2
Why would differences in morphological structure be relevant to deciding if two words form a minimal pair? The minimal pair test only refers to the phonetic segments in the words.
– user6726
Mar 10 at 23:38
2
I'm not sure I agree with this argument, but, for example, in English consonant length is generally not considered phonemic, yet if you take a word boundary, or even a morpheme boundary, you can have the same consonant twice, which results, for instance, in unaimed [ʌn'eɪmd] vs unnamed [ʌn'neɪmd], with a (potential, at least) geminate. Does this mean the analysis positing that consonant length is not phonemic in English is wrong, or does it mean that minimal pairs shouldn't be constructed across morpheme boundaries?
– LjL
Mar 11 at 0:22
1
@user6726 Rosa's vs. roses. This famous pair is not truly a minimal pair (Flemming & Johnson 2007). If English orthography marked the boundary between the stem and the plural suffix nobody would have even considered that possibility. Morphological structure is absolutely relevant because it shapes the definition of a word.
– Nardog
Mar 11 at 5:17
1
To follow Nardog, the issue is that defining what a "word" is not as clear as one may hope. So the best minimal pairs are those involving unanalyzable morphemes.
– Arnaud Fournet
Mar 11 at 10:08
@Nardog I've read the paper but it doesn't seem to say that Rosa's vs roses is not a minimal pair; I even find hints they consider it one, like "[...] a sequence of 25 minimal pairs (roses–Rosa’s, [...])". Anyway, while I think I see why you say they're not, and I've argued for that too earlier... if hypothetically Rosa's and roses were the only examples of [ɨ] vs [ə], would it be fair to say that /ɨ/ and /ə/ aren't separate phonemes in English because there is no minimal pair? I'd still say the utterances [ɹoʊzɨz] vs [ɹoʊzəz] show a meaningful distinction, indicating phonemicity.
– LjL
Mar 12 at 21:01
|
show 5 more comments
2
Why would differences in morphological structure be relevant to deciding if two words form a minimal pair? The minimal pair test only refers to the phonetic segments in the words.
– user6726
Mar 10 at 23:38
2
I'm not sure I agree with this argument, but, for example, in English consonant length is generally not considered phonemic, yet if you take a word boundary, or even a morpheme boundary, you can have the same consonant twice, which results, for instance, in unaimed [ʌn'eɪmd] vs unnamed [ʌn'neɪmd], with a (potential, at least) geminate. Does this mean the analysis positing that consonant length is not phonemic in English is wrong, or does it mean that minimal pairs shouldn't be constructed across morpheme boundaries?
– LjL
Mar 11 at 0:22
1
@user6726 Rosa's vs. roses. This famous pair is not truly a minimal pair (Flemming & Johnson 2007). If English orthography marked the boundary between the stem and the plural suffix nobody would have even considered that possibility. Morphological structure is absolutely relevant because it shapes the definition of a word.
– Nardog
Mar 11 at 5:17
1
To follow Nardog, the issue is that defining what a "word" is not as clear as one may hope. So the best minimal pairs are those involving unanalyzable morphemes.
– Arnaud Fournet
Mar 11 at 10:08
@Nardog I've read the paper but it doesn't seem to say that Rosa's vs roses is not a minimal pair; I even find hints they consider it one, like "[...] a sequence of 25 minimal pairs (roses–Rosa’s, [...])". Anyway, while I think I see why you say they're not, and I've argued for that too earlier... if hypothetically Rosa's and roses were the only examples of [ɨ] vs [ə], would it be fair to say that /ɨ/ and /ə/ aren't separate phonemes in English because there is no minimal pair? I'd still say the utterances [ɹoʊzɨz] vs [ɹoʊzəz] show a meaningful distinction, indicating phonemicity.
– LjL
Mar 12 at 21:01
2
2
Why would differences in morphological structure be relevant to deciding if two words form a minimal pair? The minimal pair test only refers to the phonetic segments in the words.
– user6726
Mar 10 at 23:38
Why would differences in morphological structure be relevant to deciding if two words form a minimal pair? The minimal pair test only refers to the phonetic segments in the words.
– user6726
Mar 10 at 23:38
2
2
I'm not sure I agree with this argument, but, for example, in English consonant length is generally not considered phonemic, yet if you take a word boundary, or even a morpheme boundary, you can have the same consonant twice, which results, for instance, in unaimed [ʌn'eɪmd] vs unnamed [ʌn'neɪmd], with a (potential, at least) geminate. Does this mean the analysis positing that consonant length is not phonemic in English is wrong, or does it mean that minimal pairs shouldn't be constructed across morpheme boundaries?
– LjL
Mar 11 at 0:22
I'm not sure I agree with this argument, but, for example, in English consonant length is generally not considered phonemic, yet if you take a word boundary, or even a morpheme boundary, you can have the same consonant twice, which results, for instance, in unaimed [ʌn'eɪmd] vs unnamed [ʌn'neɪmd], with a (potential, at least) geminate. Does this mean the analysis positing that consonant length is not phonemic in English is wrong, or does it mean that minimal pairs shouldn't be constructed across morpheme boundaries?
– LjL
Mar 11 at 0:22
1
1
@user6726 Rosa's vs. roses. This famous pair is not truly a minimal pair (Flemming & Johnson 2007). If English orthography marked the boundary between the stem and the plural suffix nobody would have even considered that possibility. Morphological structure is absolutely relevant because it shapes the definition of a word.
– Nardog
Mar 11 at 5:17
@user6726 Rosa's vs. roses. This famous pair is not truly a minimal pair (Flemming & Johnson 2007). If English orthography marked the boundary between the stem and the plural suffix nobody would have even considered that possibility. Morphological structure is absolutely relevant because it shapes the definition of a word.
– Nardog
Mar 11 at 5:17
1
1
To follow Nardog, the issue is that defining what a "word" is not as clear as one may hope. So the best minimal pairs are those involving unanalyzable morphemes.
– Arnaud Fournet
Mar 11 at 10:08
To follow Nardog, the issue is that defining what a "word" is not as clear as one may hope. So the best minimal pairs are those involving unanalyzable morphemes.
– Arnaud Fournet
Mar 11 at 10:08
@Nardog I've read the paper but it doesn't seem to say that Rosa's vs roses is not a minimal pair; I even find hints they consider it one, like "[...] a sequence of 25 minimal pairs (roses–Rosa’s, [...])". Anyway, while I think I see why you say they're not, and I've argued for that too earlier... if hypothetically Rosa's and roses were the only examples of [ɨ] vs [ə], would it be fair to say that /ɨ/ and /ə/ aren't separate phonemes in English because there is no minimal pair? I'd still say the utterances [ɹoʊzɨz] vs [ɹoʊzəz] show a meaningful distinction, indicating phonemicity.
– LjL
Mar 12 at 21:01
@Nardog I've read the paper but it doesn't seem to say that Rosa's vs roses is not a minimal pair; I even find hints they consider it one, like "[...] a sequence of 25 minimal pairs (roses–Rosa’s, [...])". Anyway, while I think I see why you say they're not, and I've argued for that too earlier... if hypothetically Rosa's and roses were the only examples of [ɨ] vs [ə], would it be fair to say that /ɨ/ and /ə/ aren't separate phonemes in English because there is no minimal pair? I'd still say the utterances [ɹoʊzɨz] vs [ɹoʊzəz] show a meaningful distinction, indicating phonemicity.
– LjL
Mar 12 at 21:01
|
show 5 more comments
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