Proving the Set of Rational Numbers is the collection of equivalence classes of ratios of integers with...
$begingroup$
I read in a textbook that the set of Rational Numbers is the collection of equivalence classes of ratios of integers with nonzero denominators. I found it interesting and tried to prove it but ended coming up with a disproof. I'm just curious as to where my logic falls through!
Assume $mathbb{Q}$ is the union of disjoint equivalence classes over ratios of integers.
Let the set of rational numbers be defined as such.
begin{equation}
mathbb{Q} = bigg{ frac{p}{q}:p,q in mathbb{Z} , q neq 0 bigg}
end{equation}
We may equivalently define the following as a definition of the rationals.
begin{equation}
mathbb{Q} = bigg{ frac{a}{b}:(a,b) in mathbb{Z} times mathbb{Z} setminus { 0 } bigg}
end{equation}
Contradiction. Equivalence classes are not possible on Cartesian products whose sets are not equal (as, at the every least, reflexivity is not possible). Therefore $mathbb{Q}$ cannot be the union of disjoint equivalence classes over ratios of integers.
equivalence-relations rational-numbers
edited Jan 4 at 8:58
Asaf Karagila♦
303k32429761
asked Jan 4 at 4:32
UmamiBoyUmamiBoy
243
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add a comment |
$begingroup$
I read in a textbook that the set of Rational Numbers is the collection of equivalence classes of ratios of integers with nonzero denominators. I found it interesting and tried to prove it but ended coming up with a disproof. I'm just curious as to where my logic falls through!
Assume $mathbb{Q}$ is the union of disjoint equivalence classes over ratios of integers.
Let the set of rational numbers be defined as such.
begin{equation}
mathbb{Q} = bigg{ frac{p}{q}:p,q in mathbb{Z} , q neq 0 bigg}
end{equation}
We may equivalently define the following as a definition of the rationals.
begin{equation}
mathbb{Q} = bigg{ frac{a}{b}:(a,b) in mathbb{Z} times mathbb{Z} setminus { 0 } bigg}
end{equation}
Contradiction. Equivalence classes are not possible on Cartesian products whose sets are not equal (as, at the every least, reflexivity is not possible). Therefore $mathbb{Q}$ cannot be the union of disjoint equivalence classes over ratios of integers.
equivalence-relations rational-numbers
edited Jan 4 at 8:58
Asaf Karagila♦
303k32429761
asked Jan 4 at 4:32
UmamiBoyUmamiBoy
243
$endgroup$
$begingroup$
What? Can you come up with a pair $(a,b)inmathbb Ztimesmathbb Zsetminus{0}$ where equivalence classes cannot be defined in the usual sense?
$endgroup$
– YiFan
Jan 4 at 6:15
add a comment |
$begingroup$
I read in a textbook that the set of Rational Numbers is the collection of equivalence classes of ratios of integers with nonzero denominators. I found it interesting and tried to prove it but ended coming up with a disproof. I'm just curious as to where my logic falls through!
Assume $mathbb{Q}$ is the union of disjoint equivalence classes over ratios of integers.
Let the set of rational numbers be defined as such.
begin{equation}
mathbb{Q} = bigg{ frac{p}{q}:p,q in mathbb{Z} , q neq 0 bigg}
end{equation}
We may equivalently define the following as a definition of the rationals.
begin{equation}
mathbb{Q} = bigg{ frac{a}{b}:(a,b) in mathbb{Z} times mathbb{Z} setminus { 0 } bigg}
end{equation}
Contradiction. Equivalence classes are not possible on Cartesian products whose sets are not equal (as, at the every least, reflexivity is not possible). Therefore $mathbb{Q}$ cannot be the union of disjoint equivalence classes over ratios of integers.
equivalence-relations rational-numbers
edited Jan 4 at 8:58
Asaf Karagila♦
303k32429761
asked Jan 4 at 4:32
UmamiBoyUmamiBoy
243
$endgroup$
I read in a textbook that the set of Rational Numbers is the collection of equivalence classes of ratios of integers with nonzero denominators. I found it interesting and tried to prove it but ended coming up with a disproof. I'm just curious as to where my logic falls through!
Assume $mathbb{Q}$ is the union of disjoint equivalence classes over ratios of integers.
Let the set of rational numbers be defined as such.
begin{equation}
mathbb{Q} = bigg{ frac{p}{q}:p,q in mathbb{Z} , q neq 0 bigg}
end{equation}
We may equivalently define the following as a definition of the rationals.
begin{equation}
mathbb{Q} = bigg{ frac{a}{b}:(a,b) in mathbb{Z} times mathbb{Z} setminus { 0 } bigg}
end{equation}
Contradiction. Equivalence classes are not possible on Cartesian products whose sets are not equal (as, at the every least, reflexivity is not possible). Therefore $mathbb{Q}$ cannot be the union of disjoint equivalence classes over ratios of integers.
equivalence-relations rational-numbers
equivalence-relations rational-numbers
edited Jan 4 at 8:58
Asaf Karagila♦
303k32429761
asked Jan 4 at 4:32
UmamiBoyUmamiBoy
243
edited Jan 4 at 8:58
Asaf Karagila♦
303k32429761
asked Jan 4 at 4:32
UmamiBoyUmamiBoy
243
edited Jan 4 at 8:58
Asaf Karagila♦
303k32429761
edited Jan 4 at 8:58
Asaf Karagila♦
303k32429761
edited Jan 4 at 8:58
Asaf Karagila♦
303k32429761
303k32429761
asked Jan 4 at 4:32
UmamiBoyUmamiBoy
243
asked Jan 4 at 4:32
UmamiBoyUmamiBoy
243
asked Jan 4 at 4:32
UmamiBoyUmamiBoy
243
243
$begingroup$
What? Can you come up with a pair $(a,b)inmathbb Ztimesmathbb Zsetminus{0}$ where equivalence classes cannot be defined in the usual sense?
$endgroup$
– YiFan
Jan 4 at 6:15
add a comment |
$begingroup$
What? Can you come up with a pair $(a,b)inmathbb Ztimesmathbb Zsetminus{0}$ where equivalence classes cannot be defined in the usual sense?
$endgroup$
– YiFan
Jan 4 at 6:15
$begingroup$
What? Can you come up with a pair $(a,b)inmathbb Ztimesmathbb Zsetminus{0}$ where equivalence classes cannot be defined in the usual sense?
$endgroup$
– YiFan
Jan 4 at 6:15
$begingroup$
What? Can you come up with a pair $(a,b)inmathbb Ztimesmathbb Zsetminus{0}$ where equivalence classes cannot be defined in the usual sense?
$endgroup$
– YiFan
Jan 4 at 6:15
add a comment |
1 Answer
1
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oldest
votes
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Equivalence classes are not possible on Cartesian products whose sets are not equal (as, at the every least, reflexivity is not possible).
Says who?
You seem to be confusing the fact that an equivalence relation on $A$ is a certain subset of $Atimes A$ with the fact that in this case $A$ itself is a cartesian product.
Here the equivalence relation is a certain subset of
$$ (mathbb Ztimes(mathbb Zsetminus{0}))times(mathbb Ztimes(mathbb Zsetminus{0})) $$
and there's nothing that prevents such a relation from being reflexive.
More precisely, the relation is
$$ { ((a,b),(p,q)) mid aq=pb } $$
which is reflexive because $((a,b),(a,b))$ is in the relation for every $ainmathbb Z$, $binmathbb Zsetminus{0}$, because $ab=ab$ is always true.
answered Jan 4 at 4:35
Henning MakholmHenning Makholm
240k17305541
$endgroup$
$begingroup$
Thank you, Having A as a cartesian product itself fixes my problem. Much appreciated.
$endgroup$
– UmamiBoy
Jan 4 at 5:09
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Equivalence classes are not possible on Cartesian products whose sets are not equal (as, at the every least, reflexivity is not possible).
Says who?
You seem to be confusing the fact that an equivalence relation on $A$ is a certain subset of $Atimes A$ with the fact that in this case $A$ itself is a cartesian product.
Here the equivalence relation is a certain subset of
$$ (mathbb Ztimes(mathbb Zsetminus{0}))times(mathbb Ztimes(mathbb Zsetminus{0})) $$
and there's nothing that prevents such a relation from being reflexive.
More precisely, the relation is
$$ { ((a,b),(p,q)) mid aq=pb } $$
which is reflexive because $((a,b),(a,b))$ is in the relation for every $ainmathbb Z$, $binmathbb Zsetminus{0}$, because $ab=ab$ is always true.
answered Jan 4 at 4:35
Henning MakholmHenning Makholm
240k17305541
$endgroup$
$begingroup$
Thank you, Having A as a cartesian product itself fixes my problem. Much appreciated.
$endgroup$
– UmamiBoy
Jan 4 at 5:09
add a comment |
$begingroup$
Equivalence classes are not possible on Cartesian products whose sets are not equal (as, at the every least, reflexivity is not possible).
Says who?
You seem to be confusing the fact that an equivalence relation on $A$ is a certain subset of $Atimes A$ with the fact that in this case $A$ itself is a cartesian product.
Here the equivalence relation is a certain subset of
$$ (mathbb Ztimes(mathbb Zsetminus{0}))times(mathbb Ztimes(mathbb Zsetminus{0})) $$
and there's nothing that prevents such a relation from being reflexive.
More precisely, the relation is
$$ { ((a,b),(p,q)) mid aq=pb } $$
which is reflexive because $((a,b),(a,b))$ is in the relation for every $ainmathbb Z$, $binmathbb Zsetminus{0}$, because $ab=ab$ is always true.
answered Jan 4 at 4:35
Henning MakholmHenning Makholm
240k17305541
$endgroup$
$begingroup$
Thank you, Having A as a cartesian product itself fixes my problem. Much appreciated.
$endgroup$
– UmamiBoy
Jan 4 at 5:09
add a comment |
$begingroup$
Equivalence classes are not possible on Cartesian products whose sets are not equal (as, at the every least, reflexivity is not possible).
Says who?
You seem to be confusing the fact that an equivalence relation on $A$ is a certain subset of $Atimes A$ with the fact that in this case $A$ itself is a cartesian product.
Here the equivalence relation is a certain subset of
$$ (mathbb Ztimes(mathbb Zsetminus{0}))times(mathbb Ztimes(mathbb Zsetminus{0})) $$
and there's nothing that prevents such a relation from being reflexive.
More precisely, the relation is
$$ { ((a,b),(p,q)) mid aq=pb } $$
which is reflexive because $((a,b),(a,b))$ is in the relation for every $ainmathbb Z$, $binmathbb Zsetminus{0}$, because $ab=ab$ is always true.
answered Jan 4 at 4:35
Henning MakholmHenning Makholm
240k17305541
$endgroup$
Equivalence classes are not possible on Cartesian products whose sets are not equal (as, at the every least, reflexivity is not possible).
Says who?
You seem to be confusing the fact that an equivalence relation on $A$ is a certain subset of $Atimes A$ with the fact that in this case $A$ itself is a cartesian product.
Here the equivalence relation is a certain subset of
$$ (mathbb Ztimes(mathbb Zsetminus{0}))times(mathbb Ztimes(mathbb Zsetminus{0})) $$
and there's nothing that prevents such a relation from being reflexive.
More precisely, the relation is
$$ { ((a,b),(p,q)) mid aq=pb } $$
which is reflexive because $((a,b),(a,b))$ is in the relation for every $ainmathbb Z$, $binmathbb Zsetminus{0}$, because $ab=ab$ is always true.
answered Jan 4 at 4:35
Henning MakholmHenning Makholm
240k17305541
answered Jan 4 at 4:35
Henning MakholmHenning Makholm
240k17305541
answered Jan 4 at 4:35
Henning MakholmHenning Makholm
240k17305541
answered Jan 4 at 4:35
Henning MakholmHenning Makholm
240k17305541
240k17305541
$begingroup$
Thank you, Having A as a cartesian product itself fixes my problem. Much appreciated.
$endgroup$
– UmamiBoy
Jan 4 at 5:09
add a comment |
$begingroup$
Thank you, Having A as a cartesian product itself fixes my problem. Much appreciated.
$endgroup$
– UmamiBoy
Jan 4 at 5:09
$begingroup$
Thank you, Having A as a cartesian product itself fixes my problem. Much appreciated.
$endgroup$
– UmamiBoy
Jan 4 at 5:09
$begingroup$
Thank you, Having A as a cartesian product itself fixes my problem. Much appreciated.
$endgroup$
– UmamiBoy
Jan 4 at 5:09
add a comment |
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$begingroup$
What? Can you come up with a pair $(a,b)inmathbb Ztimesmathbb Zsetminus{0}$ where equivalence classes cannot be defined in the usual sense?
$endgroup$
– YiFan
Jan 4 at 6:15