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If n is such a positive integer, that 8|n², then 4|n
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I'm new to the subject of discrete mathematics.
This statement is either true or false and it has to be proved. I've struggled with this exercise for quite a while, and this is what I came up with:
If 8|n² then n² is even
If n² is even then n is even
If n is positive and 4|n then n = 4k (k - any positive integer)
If n = 4k then n² = 16k²
8|16k² and 4|4k, therefore the statement is true
n|m means n divides m
So guys, can someone verify whether I proved it or not?
discrete-mathematics
asked 2 hours ago
Nojus Kudaba
141
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Nojus Kudaba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
1
down vote
favorite
I'm new to the subject of discrete mathematics.
This statement is either true or false and it has to be proved. I've struggled with this exercise for quite a while, and this is what I came up with:
If 8|n² then n² is even
If n² is even then n is even
If n is positive and 4|n then n = 4k (k - any positive integer)
If n = 4k then n² = 16k²
8|16k² and 4|4k, therefore the statement is true
n|m means n divides m
So guys, can someone verify whether I proved it or not?
discrete-mathematics
asked 2 hours ago
Nojus Kudaba
141
New contributor
Nojus Kudaba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Use the fact that if a prime $p$ divides a product of numbers, then it will divide one factor.
– Wuestenfux
2 hours ago
Your proof is wrong. You use the fact that $4|n$ in your proof. Hint is to see that $2|(n/2)^2$, so that $2|(n/2)$. I'm not sure who downvoted, but I upvoted it, I respect anyone who tries to prove something for themself
– Rumpelstiltskin
2 hours ago
What you have proven is the converse, if 4 | n then 8 | n^2
– Dark Malthorp
2 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm new to the subject of discrete mathematics.
This statement is either true or false and it has to be proved. I've struggled with this exercise for quite a while, and this is what I came up with:
If 8|n² then n² is even
If n² is even then n is even
If n is positive and 4|n then n = 4k (k - any positive integer)
If n = 4k then n² = 16k²
8|16k² and 4|4k, therefore the statement is true
n|m means n divides m
So guys, can someone verify whether I proved it or not?
discrete-mathematics
asked 2 hours ago
Nojus Kudaba
141
New contributor
Nojus Kudaba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm new to the subject of discrete mathematics.
This statement is either true or false and it has to be proved. I've struggled with this exercise for quite a while, and this is what I came up with:
If 8|n² then n² is even
If n² is even then n is even
If n is positive and 4|n then n = 4k (k - any positive integer)
If n = 4k then n² = 16k²
8|16k² and 4|4k, therefore the statement is true
n|m means n divides m
So guys, can someone verify whether I proved it or not?
discrete-mathematics
discrete-mathematics
asked 2 hours ago
Nojus Kudaba
141
New contributor
Nojus Kudaba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Nojus Kudaba
141
New contributor
Nojus Kudaba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Nojus Kudaba
141
New contributor
Nojus Kudaba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Nojus Kudaba
141
asked 2 hours ago
Nojus Kudaba
141
141
New contributor
Nojus Kudaba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Nojus Kudaba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Nojus Kudaba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Use the fact that if a prime $p$ divides a product of numbers, then it will divide one factor.
– Wuestenfux
2 hours ago
Your proof is wrong. You use the fact that $4|n$ in your proof. Hint is to see that $2|(n/2)^2$, so that $2|(n/2)$. I'm not sure who downvoted, but I upvoted it, I respect anyone who tries to prove something for themself
– Rumpelstiltskin
2 hours ago
What you have proven is the converse, if 4 | n then 8 | n^2
– Dark Malthorp
2 hours ago
add a comment |Â
Use the fact that if a prime $p$ divides a product of numbers, then it will divide one factor.
– Wuestenfux
2 hours ago
Your proof is wrong. You use the fact that $4|n$ in your proof. Hint is to see that $2|(n/2)^2$, so that $2|(n/2)$. I'm not sure who downvoted, but I upvoted it, I respect anyone who tries to prove something for themself
– Rumpelstiltskin
2 hours ago
What you have proven is the converse, if 4 | n then 8 | n^2
– Dark Malthorp
2 hours ago
Use the fact that if a prime $p$ divides a product of numbers, then it will divide one factor.
– Wuestenfux
2 hours ago
Use the fact that if a prime $p$ divides a product of numbers, then it will divide one factor.
– Wuestenfux
2 hours ago
Your proof is wrong. You use the fact that $4|n$ in your proof. Hint is to see that $2|(n/2)^2$, so that $2|(n/2)$. I'm not sure who downvoted, but I upvoted it, I respect anyone who tries to prove something for themself
– Rumpelstiltskin
2 hours ago
Your proof is wrong. You use the fact that $4|n$ in your proof. Hint is to see that $2|(n/2)^2$, so that $2|(n/2)$. I'm not sure who downvoted, but I upvoted it, I respect anyone who tries to prove something for themself
– Rumpelstiltskin
2 hours ago
What you have proven is the converse, if 4 | n then 8 | n^2
– Dark Malthorp
2 hours ago
What you have proven is the converse, if 4 | n then 8 | n^2
– Dark Malthorp
2 hours ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
2
down vote
If $k$ is the number of factors $2$ that $n$ has in its prime factorisation, then we know that $n^2$ has $2k$ such factors.
We are given that $8$ divides $n^2$ so $2k ge 3$. As $k$ is an integer this means that $k ge 2$, so indeed $4|n$.
answered 2 hours ago
Henno Brandsma
93.7k342101
add a comment |Â
up vote
2
down vote
No, you did not prove it. At no point what you wrote as something equivalent to “… and therefore $4mid n^2$â€Â.
You stated (correctly) that $n$ is even. How could $n$ then fail to be a multiple of $4$? Only if $n=2k$, whre $k$ is an odd number. But then $n^2=4k^2$ and, since $k^2$ is odd too, $8nmid4k^2$. Thereby, a contradiction is reached and so $4mid n$.
answered 2 hours ago
José Carlos Santos
122k16101186
add a comment |Â
up vote
1
down vote
Hint: the number of primes in the prime factorization of a square is even. Now $8=2^3$.
answered 2 hours ago
Michael Hoppe
9,71631532
add a comment |Â
up vote
1
down vote
Sorry, but your proof is wrong, because at a certain point you assume that $4mid n$.
Since $8mid n^2$, $n$ is even, so $n=2a$. Hence $8mid 4a^2$, which implies $2mid a^2$. Therefore $a$ is even: $a=2b$ and finally $n=4b$.
answered 2 hours ago
egreg
167k1180189
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
If $k$ is the number of factors $2$ that $n$ has in its prime factorisation, then we know that $n^2$ has $2k$ such factors.
We are given that $8$ divides $n^2$ so $2k ge 3$. As $k$ is an integer this means that $k ge 2$, so indeed $4|n$.
answered 2 hours ago
Henno Brandsma
93.7k342101
add a comment |Â
up vote
2
down vote
If $k$ is the number of factors $2$ that $n$ has in its prime factorisation, then we know that $n^2$ has $2k$ such factors.
We are given that $8$ divides $n^2$ so $2k ge 3$. As $k$ is an integer this means that $k ge 2$, so indeed $4|n$.
answered 2 hours ago
Henno Brandsma
93.7k342101
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If $k$ is the number of factors $2$ that $n$ has in its prime factorisation, then we know that $n^2$ has $2k$ such factors.
We are given that $8$ divides $n^2$ so $2k ge 3$. As $k$ is an integer this means that $k ge 2$, so indeed $4|n$.
answered 2 hours ago
Henno Brandsma
93.7k342101
If $k$ is the number of factors $2$ that $n$ has in its prime factorisation, then we know that $n^2$ has $2k$ such factors.
We are given that $8$ divides $n^2$ so $2k ge 3$. As $k$ is an integer this means that $k ge 2$, so indeed $4|n$.
answered 2 hours ago
Henno Brandsma
93.7k342101
answered 2 hours ago
Henno Brandsma
93.7k342101
answered 2 hours ago
Henno Brandsma
93.7k342101
answered 2 hours ago
Henno Brandsma
93.7k342101
93.7k342101
add a comment |Â
add a comment |Â
up vote
2
down vote
No, you did not prove it. At no point what you wrote as something equivalent to “… and therefore $4mid n^2$â€Â.
You stated (correctly) that $n$ is even. How could $n$ then fail to be a multiple of $4$? Only if $n=2k$, whre $k$ is an odd number. But then $n^2=4k^2$ and, since $k^2$ is odd too, $8nmid4k^2$. Thereby, a contradiction is reached and so $4mid n$.
answered 2 hours ago
José Carlos Santos
122k16101186
add a comment |Â
up vote
2
down vote
No, you did not prove it. At no point what you wrote as something equivalent to “… and therefore $4mid n^2$â€Â.
You stated (correctly) that $n$ is even. How could $n$ then fail to be a multiple of $4$? Only if $n=2k$, whre $k$ is an odd number. But then $n^2=4k^2$ and, since $k^2$ is odd too, $8nmid4k^2$. Thereby, a contradiction is reached and so $4mid n$.
answered 2 hours ago
José Carlos Santos
122k16101186
add a comment |Â
up vote
2
down vote
up vote
2
down vote
No, you did not prove it. At no point what you wrote as something equivalent to “… and therefore $4mid n^2$â€Â.
You stated (correctly) that $n$ is even. How could $n$ then fail to be a multiple of $4$? Only if $n=2k$, whre $k$ is an odd number. But then $n^2=4k^2$ and, since $k^2$ is odd too, $8nmid4k^2$. Thereby, a contradiction is reached and so $4mid n$.
answered 2 hours ago
José Carlos Santos
122k16101186
No, you did not prove it. At no point what you wrote as something equivalent to “… and therefore $4mid n^2$â€Â.
You stated (correctly) that $n$ is even. How could $n$ then fail to be a multiple of $4$? Only if $n=2k$, whre $k$ is an odd number. But then $n^2=4k^2$ and, since $k^2$ is odd too, $8nmid4k^2$. Thereby, a contradiction is reached and so $4mid n$.
answered 2 hours ago
José Carlos Santos
122k16101186
answered 2 hours ago
José Carlos Santos
122k16101186
answered 2 hours ago
José Carlos Santos
122k16101186
answered 2 hours ago
José Carlos Santos
122k16101186
122k16101186
add a comment |Â
add a comment |Â
up vote
1
down vote
Hint: the number of primes in the prime factorization of a square is even. Now $8=2^3$.
answered 2 hours ago
Michael Hoppe
9,71631532
add a comment |Â
up vote
1
down vote
Hint: the number of primes in the prime factorization of a square is even. Now $8=2^3$.
answered 2 hours ago
Michael Hoppe
9,71631532
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: the number of primes in the prime factorization of a square is even. Now $8=2^3$.
answered 2 hours ago
Michael Hoppe
9,71631532
Hint: the number of primes in the prime factorization of a square is even. Now $8=2^3$.
answered 2 hours ago
Michael Hoppe
9,71631532
answered 2 hours ago
Michael Hoppe
9,71631532
answered 2 hours ago
Michael Hoppe
9,71631532
answered 2 hours ago
Michael Hoppe
9,71631532
9,71631532
add a comment |Â
add a comment |Â
up vote
1
down vote
Sorry, but your proof is wrong, because at a certain point you assume that $4mid n$.
Since $8mid n^2$, $n$ is even, so $n=2a$. Hence $8mid 4a^2$, which implies $2mid a^2$. Therefore $a$ is even: $a=2b$ and finally $n=4b$.
answered 2 hours ago
egreg
167k1180189
add a comment |Â
up vote
1
down vote
Sorry, but your proof is wrong, because at a certain point you assume that $4mid n$.
Since $8mid n^2$, $n$ is even, so $n=2a$. Hence $8mid 4a^2$, which implies $2mid a^2$. Therefore $a$ is even: $a=2b$ and finally $n=4b$.
answered 2 hours ago
egreg
167k1180189
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Sorry, but your proof is wrong, because at a certain point you assume that $4mid n$.
Since $8mid n^2$, $n$ is even, so $n=2a$. Hence $8mid 4a^2$, which implies $2mid a^2$. Therefore $a$ is even: $a=2b$ and finally $n=4b$.
answered 2 hours ago
egreg
167k1180189
Sorry, but your proof is wrong, because at a certain point you assume that $4mid n$.
Since $8mid n^2$, $n$ is even, so $n=2a$. Hence $8mid 4a^2$, which implies $2mid a^2$. Therefore $a$ is even: $a=2b$ and finally $n=4b$.
answered 2 hours ago
egreg
167k1180189
answered 2 hours ago
egreg
167k1180189
answered 2 hours ago
egreg
167k1180189
answered 2 hours ago
egreg
167k1180189
167k1180189
add a comment |Â
add a comment |Â
Nojus Kudaba is a new contributor. Be nice, and check out our Code of Conduct.
Nojus Kudaba is a new contributor. Be nice, and check out our Code of Conduct.
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Use the fact that if a prime $p$ divides a product of numbers, then it will divide one factor.
– Wuestenfux
2 hours ago
Your proof is wrong. You use the fact that $4|n$ in your proof. Hint is to see that $2|(n/2)^2$, so that $2|(n/2)$. I'm not sure who downvoted, but I upvoted it, I respect anyone who tries to prove something for themself
– Rumpelstiltskin
2 hours ago
What you have proven is the converse, if 4 | n then 8 | n^2
– Dark Malthorp
2 hours ago