An error in a proof due to variable creep?
I'm working through some exam practice questions and I came across this one:
Identify the error in the following “proof.”
Let u, m, n be three integers. If u|mn and gcd(u, m) = 1, then m = ±1.
If gcd(u, m) = 1, then 1 = us + mt for some integers s, t. If u|mn, then
us = mn for some integer s. Hence, 1 = mn + mt = m(n + t), which
implies that m|1, and therefore m = ±1.
Now I think the problem here is that we cannot go from the linear transformation statement 1 = us + mt and then state that u|mn => mn = us for some integer s, because we already have a value of s in the previous statement. so we would need to define u|mn => mn = uj for some integer j.
And now the substitution doesn't work and we cannot proceed.
Does this make sense to you guys? It makes sense to me that we cannot use s in two different places here. But I'm having a hard time trying to explain why in a clear manner.
discrete-mathematics
add a comment |
I'm working through some exam practice questions and I came across this one:
Identify the error in the following “proof.”
Let u, m, n be three integers. If u|mn and gcd(u, m) = 1, then m = ±1.
If gcd(u, m) = 1, then 1 = us + mt for some integers s, t. If u|mn, then
us = mn for some integer s. Hence, 1 = mn + mt = m(n + t), which
implies that m|1, and therefore m = ±1.
Now I think the problem here is that we cannot go from the linear transformation statement 1 = us + mt and then state that u|mn => mn = us for some integer s, because we already have a value of s in the previous statement. so we would need to define u|mn => mn = uj for some integer j.
And now the substitution doesn't work and we cannot proceed.
Does this make sense to you guys? It makes sense to me that we cannot use s in two different places here. But I'm having a hard time trying to explain why in a clear manner.
discrete-mathematics
add a comment |
I'm working through some exam practice questions and I came across this one:
Identify the error in the following “proof.”
Let u, m, n be three integers. If u|mn and gcd(u, m) = 1, then m = ±1.
If gcd(u, m) = 1, then 1 = us + mt for some integers s, t. If u|mn, then
us = mn for some integer s. Hence, 1 = mn + mt = m(n + t), which
implies that m|1, and therefore m = ±1.
Now I think the problem here is that we cannot go from the linear transformation statement 1 = us + mt and then state that u|mn => mn = us for some integer s, because we already have a value of s in the previous statement. so we would need to define u|mn => mn = uj for some integer j.
And now the substitution doesn't work and we cannot proceed.
Does this make sense to you guys? It makes sense to me that we cannot use s in two different places here. But I'm having a hard time trying to explain why in a clear manner.
discrete-mathematics
I'm working through some exam practice questions and I came across this one:
Identify the error in the following “proof.”
Let u, m, n be three integers. If u|mn and gcd(u, m) = 1, then m = ±1.
If gcd(u, m) = 1, then 1 = us + mt for some integers s, t. If u|mn, then
us = mn for some integer s. Hence, 1 = mn + mt = m(n + t), which
implies that m|1, and therefore m = ±1.
Now I think the problem here is that we cannot go from the linear transformation statement 1 = us + mt and then state that u|mn => mn = us for some integer s, because we already have a value of s in the previous statement. so we would need to define u|mn => mn = uj for some integer j.
And now the substitution doesn't work and we cannot proceed.
Does this make sense to you guys? It makes sense to me that we cannot use s in two different places here. But I'm having a hard time trying to explain why in a clear manner.
discrete-mathematics
discrete-mathematics
asked Nov 26 '18 at 19:08
metisMusings
161
161
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You're right: the "proof" has defined $s$ in incompatible ways in two different places. You could explain it by analogy as follows:
We prove that $2=4$. Indeed, the number $2$ is even, so it can be expressed as $2 = 2n$ for some $n$. The number $4$ is even, so it can be expressed as $4 = 2n$ for some $n$. Hence $2 = 2n = 4$, so $2 = 4$.
Thanks, this helps a lot :)
– metisMusings
Nov 26 '18 at 19:20
add a comment |
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1 Answer
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1 Answer
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You're right: the "proof" has defined $s$ in incompatible ways in two different places. You could explain it by analogy as follows:
We prove that $2=4$. Indeed, the number $2$ is even, so it can be expressed as $2 = 2n$ for some $n$. The number $4$ is even, so it can be expressed as $4 = 2n$ for some $n$. Hence $2 = 2n = 4$, so $2 = 4$.
Thanks, this helps a lot :)
– metisMusings
Nov 26 '18 at 19:20
add a comment |
You're right: the "proof" has defined $s$ in incompatible ways in two different places. You could explain it by analogy as follows:
We prove that $2=4$. Indeed, the number $2$ is even, so it can be expressed as $2 = 2n$ for some $n$. The number $4$ is even, so it can be expressed as $4 = 2n$ for some $n$. Hence $2 = 2n = 4$, so $2 = 4$.
Thanks, this helps a lot :)
– metisMusings
Nov 26 '18 at 19:20
add a comment |
You're right: the "proof" has defined $s$ in incompatible ways in two different places. You could explain it by analogy as follows:
We prove that $2=4$. Indeed, the number $2$ is even, so it can be expressed as $2 = 2n$ for some $n$. The number $4$ is even, so it can be expressed as $4 = 2n$ for some $n$. Hence $2 = 2n = 4$, so $2 = 4$.
You're right: the "proof" has defined $s$ in incompatible ways in two different places. You could explain it by analogy as follows:
We prove that $2=4$. Indeed, the number $2$ is even, so it can be expressed as $2 = 2n$ for some $n$. The number $4$ is even, so it can be expressed as $4 = 2n$ for some $n$. Hence $2 = 2n = 4$, so $2 = 4$.
answered Nov 26 '18 at 19:16
Patrick Stevens
28.5k52874
28.5k52874
Thanks, this helps a lot :)
– metisMusings
Nov 26 '18 at 19:20
add a comment |
Thanks, this helps a lot :)
– metisMusings
Nov 26 '18 at 19:20
Thanks, this helps a lot :)
– metisMusings
Nov 26 '18 at 19:20
Thanks, this helps a lot :)
– metisMusings
Nov 26 '18 at 19:20
add a comment |
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