Question pertaining to the relationship between the GCD and LCM of 3 numbers.
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I am a high school student self-studying Number Theory and came across this question in the book Challenge and Thrill of Pre-College Mathematics (For reference, $(m,n)$ means $gcd(m,n)$ and $[m,n]$ means $text{LCM}(m,n)$):
If $m$, $n$, and $k$ are any three positive integers, prove that
$$(m,n)(m,k)(n,k)[m,n,k]^2=[m,n][m,k][n,k](m,n,k)^2$$
I was able to derive this identity by trial and error and then prove it mathematically:
$$frac{(m,n)(m,k)(n,k)[m,n,k]}{(m,n,k)}=mnk$$
And I suspect that this may be true as well:
$$frac{[m,n][m,k][n,k](m,n,k)}{[m,n,k]}=mnk$$
Which would prove the proposition. However, I am unable to prove this statement and am unsure of its truth.
Please offer a hint to how I would go about proving the second part, or in case if my assumption is incorrect than the correct method of proof.
number-theory elementary-number-theory greatest-common-divisor
asked Jan 14 at 15:08
Naman KumarNaman Kumar
8812
$endgroup$
add a comment |
$begingroup$
I am a high school student self-studying Number Theory and came across this question in the book Challenge and Thrill of Pre-College Mathematics (For reference, $(m,n)$ means $gcd(m,n)$ and $[m,n]$ means $text{LCM}(m,n)$):
If $m$, $n$, and $k$ are any three positive integers, prove that
$$(m,n)(m,k)(n,k)[m,n,k]^2=[m,n][m,k][n,k](m,n,k)^2$$
I was able to derive this identity by trial and error and then prove it mathematically:
$$frac{(m,n)(m,k)(n,k)[m,n,k]}{(m,n,k)}=mnk$$
And I suspect that this may be true as well:
$$frac{[m,n][m,k][n,k](m,n,k)}{[m,n,k]}=mnk$$
Which would prove the proposition. However, I am unable to prove this statement and am unsure of its truth.
Please offer a hint to how I would go about proving the second part, or in case if my assumption is incorrect than the correct method of proof.
number-theory elementary-number-theory greatest-common-divisor
asked Jan 14 at 15:08
Naman KumarNaman Kumar
8812
$endgroup$
add a comment |
$begingroup$
I am a high school student self-studying Number Theory and came across this question in the book Challenge and Thrill of Pre-College Mathematics (For reference, $(m,n)$ means $gcd(m,n)$ and $[m,n]$ means $text{LCM}(m,n)$):
If $m$, $n$, and $k$ are any three positive integers, prove that
$$(m,n)(m,k)(n,k)[m,n,k]^2=[m,n][m,k][n,k](m,n,k)^2$$
I was able to derive this identity by trial and error and then prove it mathematically:
$$frac{(m,n)(m,k)(n,k)[m,n,k]}{(m,n,k)}=mnk$$
And I suspect that this may be true as well:
$$frac{[m,n][m,k][n,k](m,n,k)}{[m,n,k]}=mnk$$
Which would prove the proposition. However, I am unable to prove this statement and am unsure of its truth.
Please offer a hint to how I would go about proving the second part, or in case if my assumption is incorrect than the correct method of proof.
number-theory elementary-number-theory greatest-common-divisor
asked Jan 14 at 15:08
Naman KumarNaman Kumar
8812
$endgroup$
I am a high school student self-studying Number Theory and came across this question in the book Challenge and Thrill of Pre-College Mathematics (For reference, $(m,n)$ means $gcd(m,n)$ and $[m,n]$ means $text{LCM}(m,n)$):
If $m$, $n$, and $k$ are any three positive integers, prove that
$$(m,n)(m,k)(n,k)[m,n,k]^2=[m,n][m,k][n,k](m,n,k)^2$$
I was able to derive this identity by trial and error and then prove it mathematically:
$$frac{(m,n)(m,k)(n,k)[m,n,k]}{(m,n,k)}=mnk$$
And I suspect that this may be true as well:
$$frac{[m,n][m,k][n,k](m,n,k)}{[m,n,k]}=mnk$$
Which would prove the proposition. However, I am unable to prove this statement and am unsure of its truth.
Please offer a hint to how I would go about proving the second part, or in case if my assumption is incorrect than the correct method of proof.
number-theory elementary-number-theory greatest-common-divisor
number-theory elementary-number-theory greatest-common-divisor
asked Jan 14 at 15:08
Naman KumarNaman Kumar
8812
asked Jan 14 at 15:08
Naman KumarNaman Kumar
8812
asked Jan 14 at 15:08
Naman KumarNaman Kumar
8812
asked Jan 14 at 15:08
Naman KumarNaman Kumar
8812
asked Jan 14 at 15:08
Naman KumarNaman Kumar
8812
8812
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We can use that $[m,n]cdot(m,n)=mcdot nRightarrow [m,n]=frac{mn}{(m,n)}$.
So
$$frac{[m,n][m,k][n,k](m,n,k)}{[m,n,k]}=frac{mncdot mkcdot nk cdot(m,n,k)}{(m,n)(m,k)(n,k)[m,n,k]cdot mnk} = m^2n^2k^2 cdot A $$
Where $A$ is the inverse of the term you calculated to be $mnk$.
answered Jan 14 at 15:17
YankoYanko
6,5971529
$endgroup$
$begingroup$
Wait... I think you flipped the terms around. How did $[m,n,k]$ come in the numerator and $(m,n,k)$ in the denominator? Suggested an edit.
$endgroup$
– Naman Kumar
Jan 14 at 15:33
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As far as I know, while $(m,n)[m,n]=mn$, this is not necessarily true for more than two numbers.
$endgroup$
– Naman Kumar
Jan 14 at 15:42
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@NamanKumar You're right I can't switch these two. Hold on I'll try to correct it (If I won't succeed in 5 minutes I'll delete the asnwer)
$endgroup$
– Yanko
Jan 14 at 15:42
1
$begingroup$
I think your answer is correct: we can do this except we change $[m,n]$ only for the terms with two numbers.
$endgroup$
– Naman Kumar
Jan 14 at 15:44
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@NamanKumar Yes right! I got confused because this $A$ is the inverse of $mnk$. So now it's correct :P I wish you luck with your self-study.
$endgroup$
– Yanko
Jan 14 at 15:46
add a comment |
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1 Answer
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1 Answer
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$begingroup$
We can use that $[m,n]cdot(m,n)=mcdot nRightarrow [m,n]=frac{mn}{(m,n)}$.
So
$$frac{[m,n][m,k][n,k](m,n,k)}{[m,n,k]}=frac{mncdot mkcdot nk cdot(m,n,k)}{(m,n)(m,k)(n,k)[m,n,k]cdot mnk} = m^2n^2k^2 cdot A $$
Where $A$ is the inverse of the term you calculated to be $mnk$.
answered Jan 14 at 15:17
YankoYanko
6,5971529
$endgroup$
$begingroup$
Wait... I think you flipped the terms around. How did $[m,n,k]$ come in the numerator and $(m,n,k)$ in the denominator? Suggested an edit.
$endgroup$
– Naman Kumar
Jan 14 at 15:33
$begingroup$
As far as I know, while $(m,n)[m,n]=mn$, this is not necessarily true for more than two numbers.
$endgroup$
– Naman Kumar
Jan 14 at 15:42
$begingroup$
@NamanKumar You're right I can't switch these two. Hold on I'll try to correct it (If I won't succeed in 5 minutes I'll delete the asnwer)
$endgroup$
– Yanko
Jan 14 at 15:42
1
$begingroup$
I think your answer is correct: we can do this except we change $[m,n]$ only for the terms with two numbers.
$endgroup$
– Naman Kumar
Jan 14 at 15:44
$begingroup$
@NamanKumar Yes right! I got confused because this $A$ is the inverse of $mnk$. So now it's correct :P I wish you luck with your self-study.
$endgroup$
– Yanko
Jan 14 at 15:46
add a comment |
$begingroup$
We can use that $[m,n]cdot(m,n)=mcdot nRightarrow [m,n]=frac{mn}{(m,n)}$.
So
$$frac{[m,n][m,k][n,k](m,n,k)}{[m,n,k]}=frac{mncdot mkcdot nk cdot(m,n,k)}{(m,n)(m,k)(n,k)[m,n,k]cdot mnk} = m^2n^2k^2 cdot A $$
Where $A$ is the inverse of the term you calculated to be $mnk$.
answered Jan 14 at 15:17
YankoYanko
6,5971529
$endgroup$
$begingroup$
Wait... I think you flipped the terms around. How did $[m,n,k]$ come in the numerator and $(m,n,k)$ in the denominator? Suggested an edit.
$endgroup$
– Naman Kumar
Jan 14 at 15:33
$begingroup$
As far as I know, while $(m,n)[m,n]=mn$, this is not necessarily true for more than two numbers.
$endgroup$
– Naman Kumar
Jan 14 at 15:42
$begingroup$
@NamanKumar You're right I can't switch these two. Hold on I'll try to correct it (If I won't succeed in 5 minutes I'll delete the asnwer)
$endgroup$
– Yanko
Jan 14 at 15:42
1
$begingroup$
I think your answer is correct: we can do this except we change $[m,n]$ only for the terms with two numbers.
$endgroup$
– Naman Kumar
Jan 14 at 15:44
$begingroup$
@NamanKumar Yes right! I got confused because this $A$ is the inverse of $mnk$. So now it's correct :P I wish you luck with your self-study.
$endgroup$
– Yanko
Jan 14 at 15:46
add a comment |
$begingroup$
We can use that $[m,n]cdot(m,n)=mcdot nRightarrow [m,n]=frac{mn}{(m,n)}$.
So
$$frac{[m,n][m,k][n,k](m,n,k)}{[m,n,k]}=frac{mncdot mkcdot nk cdot(m,n,k)}{(m,n)(m,k)(n,k)[m,n,k]cdot mnk} = m^2n^2k^2 cdot A $$
Where $A$ is the inverse of the term you calculated to be $mnk$.
answered Jan 14 at 15:17
YankoYanko
6,5971529
$endgroup$
We can use that $[m,n]cdot(m,n)=mcdot nRightarrow [m,n]=frac{mn}{(m,n)}$.
So
$$frac{[m,n][m,k][n,k](m,n,k)}{[m,n,k]}=frac{mncdot mkcdot nk cdot(m,n,k)}{(m,n)(m,k)(n,k)[m,n,k]cdot mnk} = m^2n^2k^2 cdot A $$
Where $A$ is the inverse of the term you calculated to be $mnk$.
answered Jan 14 at 15:17
YankoYanko
6,5971529
edited Jan 14 at 15:45
answered Jan 14 at 15:17
YankoYanko
6,5971529
answered Jan 14 at 15:17
YankoYanko
6,5971529
answered Jan 14 at 15:17
YankoYanko
6,5971529
6,5971529
$begingroup$
Wait... I think you flipped the terms around. How did $[m,n,k]$ come in the numerator and $(m,n,k)$ in the denominator? Suggested an edit.
$endgroup$
– Naman Kumar
Jan 14 at 15:33
$begingroup$
As far as I know, while $(m,n)[m,n]=mn$, this is not necessarily true for more than two numbers.
$endgroup$
– Naman Kumar
Jan 14 at 15:42
$begingroup$
@NamanKumar You're right I can't switch these two. Hold on I'll try to correct it (If I won't succeed in 5 minutes I'll delete the asnwer)
$endgroup$
– Yanko
Jan 14 at 15:42
1
$begingroup$
I think your answer is correct: we can do this except we change $[m,n]$ only for the terms with two numbers.
$endgroup$
– Naman Kumar
Jan 14 at 15:44
$begingroup$
@NamanKumar Yes right! I got confused because this $A$ is the inverse of $mnk$. So now it's correct :P I wish you luck with your self-study.
$endgroup$
– Yanko
Jan 14 at 15:46
add a comment |
$begingroup$
Wait... I think you flipped the terms around. How did $[m,n,k]$ come in the numerator and $(m,n,k)$ in the denominator? Suggested an edit.
$endgroup$
– Naman Kumar
Jan 14 at 15:33
$begingroup$
As far as I know, while $(m,n)[m,n]=mn$, this is not necessarily true for more than two numbers.
$endgroup$
– Naman Kumar
Jan 14 at 15:42
$begingroup$
@NamanKumar You're right I can't switch these two. Hold on I'll try to correct it (If I won't succeed in 5 minutes I'll delete the asnwer)
$endgroup$
– Yanko
Jan 14 at 15:42
1
$begingroup$
I think your answer is correct: we can do this except we change $[m,n]$ only for the terms with two numbers.
$endgroup$
– Naman Kumar
Jan 14 at 15:44
$begingroup$
@NamanKumar Yes right! I got confused because this $A$ is the inverse of $mnk$. So now it's correct :P I wish you luck with your self-study.
$endgroup$
– Yanko
Jan 14 at 15:46
$begingroup$
Wait... I think you flipped the terms around. How did $[m,n,k]$ come in the numerator and $(m,n,k)$ in the denominator? Suggested an edit.
$endgroup$
– Naman Kumar
Jan 14 at 15:33
$begingroup$
Wait... I think you flipped the terms around. How did $[m,n,k]$ come in the numerator and $(m,n,k)$ in the denominator? Suggested an edit.
$endgroup$
– Naman Kumar
Jan 14 at 15:33
$begingroup$
As far as I know, while $(m,n)[m,n]=mn$, this is not necessarily true for more than two numbers.
$endgroup$
– Naman Kumar
Jan 14 at 15:42
$begingroup$
As far as I know, while $(m,n)[m,n]=mn$, this is not necessarily true for more than two numbers.
$endgroup$
– Naman Kumar
Jan 14 at 15:42
$begingroup$
@NamanKumar You're right I can't switch these two. Hold on I'll try to correct it (If I won't succeed in 5 minutes I'll delete the asnwer)
$endgroup$
– Yanko
Jan 14 at 15:42
$begingroup$
@NamanKumar You're right I can't switch these two. Hold on I'll try to correct it (If I won't succeed in 5 minutes I'll delete the asnwer)
$endgroup$
– Yanko
Jan 14 at 15:42
1
1
$begingroup$
I think your answer is correct: we can do this except we change $[m,n]$ only for the terms with two numbers.
$endgroup$
– Naman Kumar
Jan 14 at 15:44
$begingroup$
I think your answer is correct: we can do this except we change $[m,n]$ only for the terms with two numbers.
$endgroup$
– Naman Kumar
Jan 14 at 15:44
$begingroup$
@NamanKumar Yes right! I got confused because this $A$ is the inverse of $mnk$. So now it's correct :P I wish you luck with your self-study.
$endgroup$
– Yanko
Jan 14 at 15:46
$begingroup$
@NamanKumar Yes right! I got confused because this $A$ is the inverse of $mnk$. So now it's correct :P I wish you luck with your self-study.
$endgroup$
– Yanko
Jan 14 at 15:46
add a comment |
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