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 $i=1,2,..,r$












1












$begingroup$



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







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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
















1












$begingroup$



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







share|cite|improve this question











$endgroup$








  • 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














1












1








1





$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 14:57









Servaes

25.6k33996




25.6k33996










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














  • 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










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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much.
    $endgroup$
    – Dima
    Jan 15 at 15:32











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much.
    $endgroup$
    – Dima
    Jan 15 at 15:32
















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much.
    $endgroup$
    – Dima
    Jan 15 at 15:32














1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 15 at 14:55









ServaesServaes

25.6k33996




25.6k33996












  • $begingroup$
    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




$begingroup$
Thank you so much.
$endgroup$
– Dima
Jan 15 at 15:32


















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