Mixing
If $c | m_i$ Prove that $c| sum_{i=1}^r u_i m_i$ where $m_i,u_i,c in mathbb{Z}$ and $i=1,2,..,r$
$begingroup$
Let $m_1,m_1,...,m_r,u_1,u_2,...,u_r,c in mathbb{Z}$, such that $c | m_i$ for each $i in lbrace 1,2,...,r rbrace$.
Prove that $c| sum_{i=1}^r u_i m_i$
My attempt:
We have $c | m_i$ for each $i in lbrace 1,2,...,r rbrace$, then
$c|u_i m_i$ for each $i in lbrace 1,2,...,r rbrace$, thus
$exists k_iin mathbb{Z}$ such that $u_i m_i=k_i c$, $forall i$
then
$sum_{i=1}^r u_i m_i=sum_{i=1}^r k_i c=c sum_{i=1}^r k_i$
and since $sum_{i=1}^r k_i in mathbb{Z}$ we got that
$c|sum_{i=1}^r u_i m_i$
Is that true please ?, thanks.
algebra-precalculus number-theory divisibility
edited Jan 15 at 14:57
Servaes
25.6k33996
asked Jan 15 at 14:45
DimaDima
736416
$endgroup$
add a comment |
$begingroup$
Let $m_1,m_1,...,m_r,u_1,u_2,...,u_r,c in mathbb{Z}$, such that $c | m_i$ for each $i in lbrace 1,2,...,r rbrace$.
Prove that $c| sum_{i=1}^r u_i m_i$
My attempt:
We have $c | m_i$ for each $i in lbrace 1,2,...,r rbrace$, then
$c|u_i m_i$ for each $i in lbrace 1,2,...,r rbrace$, thus
$exists k_iin mathbb{Z}$ such that $u_i m_i=k_i c$, $forall i$
then
$sum_{i=1}^r u_i m_i=sum_{i=1}^r k_i c=c sum_{i=1}^r k_i$
and since $sum_{i=1}^r k_i in mathbb{Z}$ we got that
$c|sum_{i=1}^r u_i m_i$
Is that true please ?, thanks.
algebra-precalculus number-theory divisibility
edited Jan 15 at 14:57
Servaes
25.6k33996
asked Jan 15 at 14:45
DimaDima
736416
$endgroup$
3
$begingroup$
This is good for me
$endgroup$
– Thomas Lesgourgues
Jan 15 at 14:55
$begingroup$
@ThomasLesgourgues Thank you so much.
$endgroup$
– Dima
Jan 15 at 15:11
add a comment |
$begingroup$
Let $m_1,m_1,...,m_r,u_1,u_2,...,u_r,c in mathbb{Z}$, such that $c | m_i$ for each $i in lbrace 1,2,...,r rbrace$.
Prove that $c| sum_{i=1}^r u_i m_i$
My attempt:
We have $c | m_i$ for each $i in lbrace 1,2,...,r rbrace$, then
$c|u_i m_i$ for each $i in lbrace 1,2,...,r rbrace$, thus
$exists k_iin mathbb{Z}$ such that $u_i m_i=k_i c$, $forall i$
then
$sum_{i=1}^r u_i m_i=sum_{i=1}^r k_i c=c sum_{i=1}^r k_i$
and since $sum_{i=1}^r k_i in mathbb{Z}$ we got that
$c|sum_{i=1}^r u_i m_i$
Is that true please ?, thanks.
algebra-precalculus number-theory divisibility
edited Jan 15 at 14:57
Servaes
25.6k33996
asked Jan 15 at 14:45
DimaDima
736416
$endgroup$
Let $m_1,m_1,...,m_r,u_1,u_2,...,u_r,c in mathbb{Z}$, such that $c | m_i$ for each $i in lbrace 1,2,...,r rbrace$.
Prove that $c| sum_{i=1}^r u_i m_i$
My attempt:
We have $c | m_i$ for each $i in lbrace 1,2,...,r rbrace$, then
$c|u_i m_i$ for each $i in lbrace 1,2,...,r rbrace$, thus
$exists k_iin mathbb{Z}$ such that $u_i m_i=k_i c$, $forall i$
then
$sum_{i=1}^r u_i m_i=sum_{i=1}^r k_i c=c sum_{i=1}^r k_i$
and since $sum_{i=1}^r k_i in mathbb{Z}$ we got that
$c|sum_{i=1}^r u_i m_i$
Is that true please ?, thanks.
algebra-precalculus number-theory divisibility
algebra-precalculus number-theory divisibility
edited Jan 15 at 14:57
Servaes
25.6k33996
asked Jan 15 at 14:45
DimaDima
736416
edited Jan 15 at 14:57
Servaes
25.6k33996
asked Jan 15 at 14:45
DimaDima
736416
edited Jan 15 at 14:57
Servaes
25.6k33996
edited Jan 15 at 14:57
Servaes
25.6k33996
edited Jan 15 at 14:57
Servaes
25.6k33996
25.6k33996
asked Jan 15 at 14:45
DimaDima
736416
asked Jan 15 at 14:45
DimaDima
736416
asked Jan 15 at 14:45
DimaDima
736416
736416
3
$begingroup$
This is good for me
$endgroup$
– Thomas Lesgourgues
Jan 15 at 14:55
$begingroup$
@ThomasLesgourgues Thank you so much.
$endgroup$
– Dima
Jan 15 at 15:11
add a comment |
3
$begingroup$
This is good for me
$endgroup$
– Thomas Lesgourgues
Jan 15 at 14:55
$begingroup$
@ThomasLesgourgues Thank you so much.
$endgroup$
– Dima
Jan 15 at 15:11
3
3
$begingroup$
This is good for me
$endgroup$
– Thomas Lesgourgues
Jan 15 at 14:55
$begingroup$
This is good for me
$endgroup$
– Thomas Lesgourgues
Jan 15 at 14:55
$begingroup$
@ThomasLesgourgues Thank you so much.
$endgroup$
– Dima
Jan 15 at 15:11
$begingroup$
@ThomasLesgourgues Thank you so much.
$endgroup$
– Dima
Jan 15 at 15:11
add a comment |
1 Answer
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$begingroup$
Your proof is entirely correct. You might also use the fact that $cmid m_i$ to deduce that $m_i=n_ic$ and hence that $u_im_i=u_in_ic$, to reach the same conclusion.
answered Jan 15 at 14:55
ServaesServaes
25.6k33996
$endgroup$
$begingroup$
Thank you so much.
$endgroup$
– Dima
Jan 15 at 15:32
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your proof is entirely correct. You might also use the fact that $cmid m_i$ to deduce that $m_i=n_ic$ and hence that $u_im_i=u_in_ic$, to reach the same conclusion.
answered Jan 15 at 14:55
ServaesServaes
25.6k33996
$endgroup$
$begingroup$
Thank you so much.
$endgroup$
– Dima
Jan 15 at 15:32
add a comment |
$begingroup$
Your proof is entirely correct. You might also use the fact that $cmid m_i$ to deduce that $m_i=n_ic$ and hence that $u_im_i=u_in_ic$, to reach the same conclusion.
answered Jan 15 at 14:55
ServaesServaes
25.6k33996
$endgroup$
$begingroup$
Thank you so much.
$endgroup$
– Dima
Jan 15 at 15:32
add a comment |
$begingroup$
Your proof is entirely correct. You might also use the fact that $cmid m_i$ to deduce that $m_i=n_ic$ and hence that $u_im_i=u_in_ic$, to reach the same conclusion.
answered Jan 15 at 14:55
ServaesServaes
25.6k33996
$endgroup$
Your proof is entirely correct. You might also use the fact that $cmid m_i$ to deduce that $m_i=n_ic$ and hence that $u_im_i=u_in_ic$, to reach the same conclusion.
answered Jan 15 at 14:55
ServaesServaes
25.6k33996
answered Jan 15 at 14:55
ServaesServaes
25.6k33996
answered Jan 15 at 14:55
ServaesServaes
25.6k33996
answered Jan 15 at 14:55
ServaesServaes
25.6k33996
25.6k33996
$begingroup$
Thank you so much.
$endgroup$
– Dima
Jan 15 at 15:32
add a comment |
$begingroup$
Thank you so much.
$endgroup$
– Dima
Jan 15 at 15:32
$begingroup$
Thank you so much.
$endgroup$
– Dima
Jan 15 at 15:32
$begingroup$
Thank you so much.
$endgroup$
– Dima
Jan 15 at 15:32
add a comment |
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3
$begingroup$
This is good for me
$endgroup$
– Thomas Lesgourgues
Jan 15 at 14:55
$begingroup$
@ThomasLesgourgues Thank you so much.
$endgroup$
– Dima
Jan 15 at 15:11