Mountain of Books
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I need help with proving that a set of nonzero rational numbers is a commutative group under multiplication using variables.
I need help for a proof!
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Steven G
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What have tried? Do you know the definition of a commutative group? Do you know what the set of non-zero rational numbers are?
This is a help site not a homework site. You need to solve your own problem with our help.
Start off by knowing the four properties that are required for a group.
Hint: The first one is closure. Let x and y be non-zero rational numbers. You need to argue why xy is a non-zero rational number.
Then you go on to show the 2nd property of a group.
Post back showing your work.
This is a help site not a homework site. You need to solve your own problem with our help.
Start off by knowing the four properties that are required for a group.
Hint: The first one is closure. Let x and y be non-zero rational numbers. You need to argue why xy is a non-zero rational number.
Then you go on to show the 2nd property of a group.
Post back showing your work.
pka
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The set \(\mathbb{Q}^*=\mathbb{Q}\setminus\{0\}\) is the set of non-zero rationals.
Now clearly \(\mathbb{Q}^*\) is a commutative group with unity. Why is that?
What more do you need?
HallsofIvy
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To prove that a set with a given binary operation is a "commutative group" you need to show
1) a(bc)= (ab)c (associative law)
2) there exist an "identity", e, such that ae= a for all a (existence of identity)
3) every element, a, has an inverse, \(\displaystyle a^{-1}\) such that \(\displaystyle a(a^{-1})= e\) (existence of inverses)
4) for any elements, a, b, ab = ba (commutative law)
(A general "group", not necessarily commutative, only has to satisfy the first three.)
Here the members of the group are all non zero rational numbers. Any such number can be written as "\(\displaystyle \frac{m}{n}\)" where m and n are integers and m is not 0.
The associative law requires that you show that \(\displaystyle \left[\left(\frac{m}{n}\right)\left(\frac{p}{q}\right)\right]\left(\frac{r}{s}\right)= \left(\frac{m}{n}\right)\left[\left(\frac{p}{q}\right)\left(\frac{r}{s}\right)\right]\). That is, \(\displaystyle \frac{(mp)r}{(nq)s}= \frac{m(pr)}{n(qs)}\). I presume you are allowed to use the fact that multiplication of integers is associative.
The identity element and the inverse of \(\displaystyle \frac{a}{b}\) should be obvious (It is important that 0 iss not included- 0 does not have a multiplicative inverse).
To prove the commutative law, you need to show that \(\displaystyle \left(\frac{a}{b}\right)\left(\frac{p}{q}\right)= \left(\frac{p}{q}\right)\left(\frac{a}{b}\right)\). Again use the fact that multiplication of integers is commutative.
1) a(bc)= (ab)c (associative law)
2) there exist an "identity", e, such that ae= a for all a (existence of identity)
3) every element, a, has an inverse, \(\displaystyle a^{-1}\) such that \(\displaystyle a(a^{-1})= e\) (existence of inverses)
4) for any elements, a, b, ab = ba (commutative law)
(A general "group", not necessarily commutative, only has to satisfy the first three.)
Here the members of the group are all non zero rational numbers. Any such number can be written as "\(\displaystyle \frac{m}{n}\)" where m and n are integers and m is not 0.
The associative law requires that you show that \(\displaystyle \left[\left(\frac{m}{n}\right)\left(\frac{p}{q}\right)\right]\left(\frac{r}{s}\right)= \left(\frac{m}{n}\right)\left[\left(\frac{p}{q}\right)\left(\frac{r}{s}\right)\right]\). That is, \(\displaystyle \frac{(mp)r}{(nq)s}= \frac{m(pr)}{n(qs)}\). I presume you are allowed to use the fact that multiplication of integers is associative.
The identity element and the inverse of \(\displaystyle \frac{a}{b}\) should be obvious (It is important that 0 iss not included- 0 does not have a multiplicative inverse).
To prove the commutative law, you need to show that \(\displaystyle \left(\frac{a}{b}\right)\left(\frac{p}{q}\right)= \left(\frac{p}{q}\right)\left(\frac{a}{b}\right)\). Again use the fact that multiplication of integers is commutative.
Steven G
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Prof Ivy, It has been a long time since I studied Algebra and I have a simple question for you. Are you saying that you do not need to show closure because a binary operation must be closed? The more I think about it the less sense it makes.