Determine whether $*$ is associative, where $*$ is defined on $Bbb Z^+$ by $a*b=2^{ab}$.
I am doing this group theory question:
I have already proven that * is commutative, however, I'm I bit confused about proving for associativity.
I used three variables a, b and c and said: RTF whether a*(bc) = (ab)*c
For the right hand side of the equation I got: 2^(2^(bc) a)
For the left hand side of the equation I got: 2^(2^(ab) c)
I know these two appear to be different, but my question is, is there away to simplify what is there, because I'm unsure if they are actually the same or not.
group-theory
edited Nov 28 '18 at 18:21
Shaun
8,810113680
asked Jun 15 '16 at 4:04
Genny Murphy
325
add a comment |
I am doing this group theory question:
I have already proven that * is commutative, however, I'm I bit confused about proving for associativity.
I used three variables a, b and c and said: RTF whether a*(bc) = (ab)*c
For the right hand side of the equation I got: 2^(2^(bc) a)
For the left hand side of the equation I got: 2^(2^(ab) c)
I know these two appear to be different, but my question is, is there away to simplify what is there, because I'm unsure if they are actually the same or not.
group-theory
edited Nov 28 '18 at 18:21
Shaun
8,810113680
asked Jun 15 '16 at 4:04
Genny Murphy
325
Try some small $a,b,c$, say $b=c=1$. You can decide about $a$.
– André Nicolas
Jun 15 '16 at 4:11
add a comment |
I am doing this group theory question:
I have already proven that * is commutative, however, I'm I bit confused about proving for associativity.
I used three variables a, b and c and said: RTF whether a*(bc) = (ab)*c
For the right hand side of the equation I got: 2^(2^(bc) a)
For the left hand side of the equation I got: 2^(2^(ab) c)
I know these two appear to be different, but my question is, is there away to simplify what is there, because I'm unsure if they are actually the same or not.
group-theory
edited Nov 28 '18 at 18:21
Shaun
8,810113680
asked Jun 15 '16 at 4:04
Genny Murphy
325
I am doing this group theory question:
I have already proven that * is commutative, however, I'm I bit confused about proving for associativity.
I used three variables a, b and c and said: RTF whether a*(bc) = (ab)*c
For the right hand side of the equation I got: 2^(2^(bc) a)
For the left hand side of the equation I got: 2^(2^(ab) c)
I know these two appear to be different, but my question is, is there away to simplify what is there, because I'm unsure if they are actually the same or not.
group-theory
group-theory
edited Nov 28 '18 at 18:21
Shaun
8,810113680
asked Jun 15 '16 at 4:04
Genny Murphy
325
edited Nov 28 '18 at 18:21
Shaun
8,810113680
asked Jun 15 '16 at 4:04
Genny Murphy
325
edited Nov 28 '18 at 18:21
Shaun
8,810113680
edited Nov 28 '18 at 18:21
Shaun
8,810113680
edited Nov 28 '18 at 18:21
Shaun
8,810113680
8,810113680
asked Jun 15 '16 at 4:04
Genny Murphy
325
asked Jun 15 '16 at 4:04
Genny Murphy
325
asked Jun 15 '16 at 4:04
Genny Murphy
325
325
Try some small $a,b,c$, say $b=c=1$. You can decide about $a$.
– André Nicolas
Jun 15 '16 at 4:11
add a comment |
Try some small $a,b,c$, say $b=c=1$. You can decide about $a$.
– André Nicolas
Jun 15 '16 at 4:11
Try some small $a,b,c$, say $b=c=1$. You can decide about $a$.
– André Nicolas
Jun 15 '16 at 4:11
Try some small $a,b,c$, say $b=c=1$. You can decide about $a$.
– André Nicolas
Jun 15 '16 at 4:11
add a comment |
2 Answers
2
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As you say, $(a*b)*c = 2^{(a*b)c} = 2^{2^{ab}c}$ vs. $a*(b*c) = 2^{a(b*c)} = 2^{a2^{bc}}$. For these two to be equal, then we would need:
$$
c2^{ab} = a2^{bc}
$$
These are clearly not equal for all $a$, $b$, $c$. Simply choose $a$ and $c$ to be relatively prime and then there is no way for these to be equal (in fact, I suspect the only way for these to be equal is for $a = c$).
answered Jun 15 '16 at 4:25
Jared
5,19311116
add a comment |
You can easily find a counterexample to these. Just take the numbers $2,3,5$. We have that
begin{eqnarray*}
(2*3)*5 & = & (2^6)*5 ;; =;; 2^{64times 5} ;; =;; 2^{320} \
2*(3*5) & = & 2*(2^{15}) ;; =;; 2^{2 times 32,768} ;; =;; 2^{65,536}
end{eqnarray*}
which are quite different from one another.
answered Jun 15 '16 at 4:09
Mnifldz
6,81311534
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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As you say, $(a*b)*c = 2^{(a*b)c} = 2^{2^{ab}c}$ vs. $a*(b*c) = 2^{a(b*c)} = 2^{a2^{bc}}$. For these two to be equal, then we would need:
$$
c2^{ab} = a2^{bc}
$$
These are clearly not equal for all $a$, $b$, $c$. Simply choose $a$ and $c$ to be relatively prime and then there is no way for these to be equal (in fact, I suspect the only way for these to be equal is for $a = c$).
answered Jun 15 '16 at 4:25
Jared
5,19311116
add a comment |
As you say, $(a*b)*c = 2^{(a*b)c} = 2^{2^{ab}c}$ vs. $a*(b*c) = 2^{a(b*c)} = 2^{a2^{bc}}$. For these two to be equal, then we would need:
$$
c2^{ab} = a2^{bc}
$$
These are clearly not equal for all $a$, $b$, $c$. Simply choose $a$ and $c$ to be relatively prime and then there is no way for these to be equal (in fact, I suspect the only way for these to be equal is for $a = c$).
answered Jun 15 '16 at 4:25
Jared
5,19311116
add a comment |
As you say, $(a*b)*c = 2^{(a*b)c} = 2^{2^{ab}c}$ vs. $a*(b*c) = 2^{a(b*c)} = 2^{a2^{bc}}$. For these two to be equal, then we would need:
$$
c2^{ab} = a2^{bc}
$$
These are clearly not equal for all $a$, $b$, $c$. Simply choose $a$ and $c$ to be relatively prime and then there is no way for these to be equal (in fact, I suspect the only way for these to be equal is for $a = c$).
answered Jun 15 '16 at 4:25
Jared
5,19311116
As you say, $(a*b)*c = 2^{(a*b)c} = 2^{2^{ab}c}$ vs. $a*(b*c) = 2^{a(b*c)} = 2^{a2^{bc}}$. For these two to be equal, then we would need:
$$
c2^{ab} = a2^{bc}
$$
These are clearly not equal for all $a$, $b$, $c$. Simply choose $a$ and $c$ to be relatively prime and then there is no way for these to be equal (in fact, I suspect the only way for these to be equal is for $a = c$).
answered Jun 15 '16 at 4:25
Jared
5,19311116
answered Jun 15 '16 at 4:25
Jared
5,19311116
answered Jun 15 '16 at 4:25
Jared
5,19311116
answered Jun 15 '16 at 4:25
Jared
5,19311116
5,19311116
add a comment |
add a comment |
You can easily find a counterexample to these. Just take the numbers $2,3,5$. We have that
begin{eqnarray*}
(2*3)*5 & = & (2^6)*5 ;; =;; 2^{64times 5} ;; =;; 2^{320} \
2*(3*5) & = & 2*(2^{15}) ;; =;; 2^{2 times 32,768} ;; =;; 2^{65,536}
end{eqnarray*}
which are quite different from one another.
answered Jun 15 '16 at 4:09
Mnifldz
6,81311534
add a comment |
You can easily find a counterexample to these. Just take the numbers $2,3,5$. We have that
begin{eqnarray*}
(2*3)*5 & = & (2^6)*5 ;; =;; 2^{64times 5} ;; =;; 2^{320} \
2*(3*5) & = & 2*(2^{15}) ;; =;; 2^{2 times 32,768} ;; =;; 2^{65,536}
end{eqnarray*}
which are quite different from one another.
answered Jun 15 '16 at 4:09
Mnifldz
6,81311534
add a comment |
You can easily find a counterexample to these. Just take the numbers $2,3,5$. We have that
begin{eqnarray*}
(2*3)*5 & = & (2^6)*5 ;; =;; 2^{64times 5} ;; =;; 2^{320} \
2*(3*5) & = & 2*(2^{15}) ;; =;; 2^{2 times 32,768} ;; =;; 2^{65,536}
end{eqnarray*}
which are quite different from one another.
answered Jun 15 '16 at 4:09
Mnifldz
6,81311534
You can easily find a counterexample to these. Just take the numbers $2,3,5$. We have that
begin{eqnarray*}
(2*3)*5 & = & (2^6)*5 ;; =;; 2^{64times 5} ;; =;; 2^{320} \
2*(3*5) & = & 2*(2^{15}) ;; =;; 2^{2 times 32,768} ;; =;; 2^{65,536}
end{eqnarray*}
which are quite different from one another.
answered Jun 15 '16 at 4:09
Mnifldz
6,81311534
answered Jun 15 '16 at 4:09
Mnifldz
6,81311534
answered Jun 15 '16 at 4:09
Mnifldz
6,81311534
answered Jun 15 '16 at 4:09
Mnifldz
6,81311534
6,81311534
add a comment |
add a comment |
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Try some small $a,b,c$, say $b=c=1$. You can decide about $a$.
– André Nicolas
Jun 15 '16 at 4:11