Sequence of Number System Construction
1
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After constructing the naturals, why construct integers before rationals? Is there a historical explanation? Couldn't ordered pairs of fractions constructed from the naturals be used to represent equivalence classes capable of annihilating each other through addition? Would there be downstream implications for constructions of the reals or complex numbers?
integers rational-numbers foundations
asked Dec 12 '18 at 23:33
bblohowiakbblohowiak
1099
$endgroup$
add a comment |
1
$begingroup$
After constructing the naturals, why construct integers before rationals? Is there a historical explanation? Couldn't ordered pairs of fractions constructed from the naturals be used to represent equivalence classes capable of annihilating each other through addition? Would there be downstream implications for constructions of the reals or complex numbers?
integers rational-numbers foundations
asked Dec 12 '18 at 23:33
bblohowiakbblohowiak
1099
$endgroup$
$begingroup$
Depends on what you mean by "rationals." Since the $mathbb{Q}$ we know is a field, you'd need additive inverses, and I don't know if there's a clear way to get those without just defining the integers implicitly anyways.
$endgroup$
– platty
Dec 12 '18 at 23:35
1
$begingroup$
If you can define rationals first, then go for it! You could easily define the non-negative rationals first without incident, then extend these to negative rationals too, and define the integers as a subset. There's no "right" way to do this (only many wrong ways). Any equivalent construction will work just fine.
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:47
1
$begingroup$
Yes, I think you can construct the nonnegative rationals from the naturals as $(mathbb{N} times mathbb{N^+}) / sim$ where $(a, b) sim (c, d)$ if $ad = bc$. You can check $~$ is an equivalence relation as before.
$endgroup$
– Alex Vong
Dec 12 '18 at 23:54
add a comment |
1
1
1
$begingroup$
After constructing the naturals, why construct integers before rationals? Is there a historical explanation? Couldn't ordered pairs of fractions constructed from the naturals be used to represent equivalence classes capable of annihilating each other through addition? Would there be downstream implications for constructions of the reals or complex numbers?
integers rational-numbers foundations
asked Dec 12 '18 at 23:33
bblohowiakbblohowiak
1099
$endgroup$
After constructing the naturals, why construct integers before rationals? Is there a historical explanation? Couldn't ordered pairs of fractions constructed from the naturals be used to represent equivalence classes capable of annihilating each other through addition? Would there be downstream implications for constructions of the reals or complex numbers?
integers rational-numbers foundations
integers rational-numbers foundations
asked Dec 12 '18 at 23:33
bblohowiakbblohowiak
1099
asked Dec 12 '18 at 23:33
bblohowiakbblohowiak
1099
asked Dec 12 '18 at 23:33
bblohowiakbblohowiak
1099
asked Dec 12 '18 at 23:33
bblohowiakbblohowiak
1099
asked Dec 12 '18 at 23:33
bblohowiakbblohowiak
1099
1099
$begingroup$
Depends on what you mean by "rationals." Since the $mathbb{Q}$ we know is a field, you'd need additive inverses, and I don't know if there's a clear way to get those without just defining the integers implicitly anyways.
$endgroup$
– platty
Dec 12 '18 at 23:35
1
$begingroup$
If you can define rationals first, then go for it! You could easily define the non-negative rationals first without incident, then extend these to negative rationals too, and define the integers as a subset. There's no "right" way to do this (only many wrong ways). Any equivalent construction will work just fine.
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:47
1
$begingroup$
Yes, I think you can construct the nonnegative rationals from the naturals as $(mathbb{N} times mathbb{N^+}) / sim$ where $(a, b) sim (c, d)$ if $ad = bc$. You can check $~$ is an equivalence relation as before.
$endgroup$
– Alex Vong
Dec 12 '18 at 23:54
add a comment |
$begingroup$
Depends on what you mean by "rationals." Since the $mathbb{Q}$ we know is a field, you'd need additive inverses, and I don't know if there's a clear way to get those without just defining the integers implicitly anyways.
$endgroup$
– platty
Dec 12 '18 at 23:35
1
$begingroup$
If you can define rationals first, then go for it! You could easily define the non-negative rationals first without incident, then extend these to negative rationals too, and define the integers as a subset. There's no "right" way to do this (only many wrong ways). Any equivalent construction will work just fine.
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:47
1
$begingroup$
Yes, I think you can construct the nonnegative rationals from the naturals as $(mathbb{N} times mathbb{N^+}) / sim$ where $(a, b) sim (c, d)$ if $ad = bc$. You can check $~$ is an equivalence relation as before.
$endgroup$
– Alex Vong
Dec 12 '18 at 23:54
$begingroup$
Depends on what you mean by "rationals." Since the $mathbb{Q}$ we know is a field, you'd need additive inverses, and I don't know if there's a clear way to get those without just defining the integers implicitly anyways.
$endgroup$
– platty
Dec 12 '18 at 23:35
$begingroup$
Depends on what you mean by "rationals." Since the $mathbb{Q}$ we know is a field, you'd need additive inverses, and I don't know if there's a clear way to get those without just defining the integers implicitly anyways.
$endgroup$
– platty
Dec 12 '18 at 23:35
1
1
$begingroup$
If you can define rationals first, then go for it! You could easily define the non-negative rationals first without incident, then extend these to negative rationals too, and define the integers as a subset. There's no "right" way to do this (only many wrong ways). Any equivalent construction will work just fine.
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:47
$begingroup$
If you can define rationals first, then go for it! You could easily define the non-negative rationals first without incident, then extend these to negative rationals too, and define the integers as a subset. There's no "right" way to do this (only many wrong ways). Any equivalent construction will work just fine.
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:47
1
1
$begingroup$
Yes, I think you can construct the nonnegative rationals from the naturals as $(mathbb{N} times mathbb{N^+}) / sim$ where $(a, b) sim (c, d)$ if $ad = bc$. You can check $~$ is an equivalence relation as before.
$endgroup$
– Alex Vong
Dec 12 '18 at 23:54
$begingroup$
Yes, I think you can construct the nonnegative rationals from the naturals as $(mathbb{N} times mathbb{N^+}) / sim$ where $(a, b) sim (c, d)$ if $ad = bc$. You can check $~$ is an equivalence relation as before.
$endgroup$
– Alex Vong
Dec 12 '18 at 23:54
add a comment |
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$begingroup$
Depends on what you mean by "rationals." Since the $mathbb{Q}$ we know is a field, you'd need additive inverses, and I don't know if there's a clear way to get those without just defining the integers implicitly anyways.
$endgroup$
– platty
Dec 12 '18 at 23:35
1
$begingroup$
If you can define rationals first, then go for it! You could easily define the non-negative rationals first without incident, then extend these to negative rationals too, and define the integers as a subset. There's no "right" way to do this (only many wrong ways). Any equivalent construction will work just fine.
$endgroup$
– Theo Bendit
Dec 12 '18 at 23:47
1
$begingroup$
Yes, I think you can construct the nonnegative rationals from the naturals as $(mathbb{N} times mathbb{N^+}) / sim$ where $(a, b) sim (c, d)$ if $ad = bc$. You can check $~$ is an equivalence relation as before.
$endgroup$
– Alex Vong
Dec 12 '18 at 23:54