Elementary Number Theory - Divisibility [duplicate]
This question already has an answer here:
Prove that If $m'$ is a common multiple of $s$ and $t$, then $m | m'$. Here $m$ is the LCM of $s$ and $t$. [duplicate]
2 answers
If $a mid r$ and $b mid r$ show that $operatorname{lcm}(a,b) mid r$ by using only elementary properties of numbers.
elementary-number-theory
edited Nov 28 '18 at 11:23
amWhy
192k28224439
asked Nov 28 '18 at 10:59
prashant sharma
576
marked as duplicate by Bill Dubuque elementary-number-theory
Users with the elementary-number-theory badge can single-handedly close elementary-number-theory questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 28 '18 at 15:17
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Prove that If $m'$ is a common multiple of $s$ and $t$, then $m | m'$. Here $m$ is the LCM of $s$ and $t$. [duplicate]
2 answers
If $a mid r$ and $b mid r$ show that $operatorname{lcm}(a,b) mid r$ by using only elementary properties of numbers.
elementary-number-theory
edited Nov 28 '18 at 11:23
amWhy
192k28224439
asked Nov 28 '18 at 10:59
prashant sharma
576
marked as duplicate by Bill Dubuque elementary-number-theory
Users with the elementary-number-theory badge can single-handedly close elementary-number-theory questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 28 '18 at 15:17
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Prove that If $m'$ is a common multiple of $s$ and $t$, then $m | m'$. Here $m$ is the LCM of $s$ and $t$. [duplicate]
2 answers
If $a mid r$ and $b mid r$ show that $operatorname{lcm}(a,b) mid r$ by using only elementary properties of numbers.
elementary-number-theory
edited Nov 28 '18 at 11:23
amWhy
192k28224439
asked Nov 28 '18 at 10:59
prashant sharma
576
This question already has an answer here:
Prove that If $m'$ is a common multiple of $s$ and $t$, then $m | m'$. Here $m$ is the LCM of $s$ and $t$. [duplicate]
2 answers
If $a mid r$ and $b mid r$ show that $operatorname{lcm}(a,b) mid r$ by using only elementary properties of numbers.
This question already has an answer here:
Prove that If $m'$ is a common multiple of $s$ and $t$, then $m | m'$. Here $m$ is the LCM of $s$ and $t$. [duplicate]
2 answers
elementary-number-theory
elementary-number-theory
edited Nov 28 '18 at 11:23
amWhy
192k28224439
asked Nov 28 '18 at 10:59
prashant sharma
576
edited Nov 28 '18 at 11:23
amWhy
192k28224439
asked Nov 28 '18 at 10:59
prashant sharma
576
edited Nov 28 '18 at 11:23
amWhy
192k28224439
edited Nov 28 '18 at 11:23
amWhy
192k28224439
edited Nov 28 '18 at 11:23
amWhy
192k28224439
192k28224439
asked Nov 28 '18 at 10:59
prashant sharma
576
asked Nov 28 '18 at 10:59
prashant sharma
576
asked Nov 28 '18 at 10:59
prashant sharma
576
576
marked as duplicate by Bill Dubuque elementary-number-theory
Users with the elementary-number-theory badge can single-handedly close elementary-number-theory questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 28 '18 at 15:17
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Bill Dubuque elementary-number-theory
Users with the elementary-number-theory badge can single-handedly close elementary-number-theory questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 28 '18 at 15:17
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
Let $m=mbox{lcm}(a,b).$ By division algorithm $r=qm+s$ with $0leq s < m.$ Then $a mid r = qm+s$. Since $a$ also divides $m$, we have $a mid s$. Similarly $bmid s.$ Therefore $s$ is a common multiple of $a$ and $b$. But $m$ is the least common multiple and $s$ is less than $m$. So $s=0$.
answered Nov 28 '18 at 11:30
B. Goddard
18.4k21340
add a comment |
$text{lcm}(a,b)=q$. By division algorithm, $r=mq+cquad 0leq c<q.$ We have $a|r,a|qimplies a|c$ and $b|r,b|qimplies b|c$. Since $c$ is a common multiple of $a$ and $b$ with $:c<q$. So $c=0$ and $q|r$.
answered Nov 28 '18 at 11:17
Yadati Kiran
1,694619
No. $amid c$ and $bmid c$ does not imply that $abmid c$.
– B. Goddard
Nov 28 '18 at 11:25
@B.Goddard: It was a huge oversight. Thanks.
– Yadati Kiran
Nov 28 '18 at 11:31
add a comment |
Consider the following theorem:
If $m=operatorname{lcm}(a,b)$ then $m|r$ for any common multiple $r$ of $a,b$.
You can, roughly speaking, say that, since $a|r,b|r$, then $r$ is a common multiple of $a,b$. But each common multiple of $a,b$ is a multiple of $operatorname{lcm}(a,b)$. Therefore $r=qoperatorname{lcm}(a,b)$. And the proof completes.
The proof of the theorem can be inferred from @B. Goddard answer.
answered Nov 28 '18 at 13:10
Maged Saeed
8521417
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $m=mbox{lcm}(a,b).$ By division algorithm $r=qm+s$ with $0leq s < m.$ Then $a mid r = qm+s$. Since $a$ also divides $m$, we have $a mid s$. Similarly $bmid s.$ Therefore $s$ is a common multiple of $a$ and $b$. But $m$ is the least common multiple and $s$ is less than $m$. So $s=0$.
answered Nov 28 '18 at 11:30
B. Goddard
18.4k21340
add a comment |
Let $m=mbox{lcm}(a,b).$ By division algorithm $r=qm+s$ with $0leq s < m.$ Then $a mid r = qm+s$. Since $a$ also divides $m$, we have $a mid s$. Similarly $bmid s.$ Therefore $s$ is a common multiple of $a$ and $b$. But $m$ is the least common multiple and $s$ is less than $m$. So $s=0$.
answered Nov 28 '18 at 11:30
B. Goddard
18.4k21340
add a comment |
Let $m=mbox{lcm}(a,b).$ By division algorithm $r=qm+s$ with $0leq s < m.$ Then $a mid r = qm+s$. Since $a$ also divides $m$, we have $a mid s$. Similarly $bmid s.$ Therefore $s$ is a common multiple of $a$ and $b$. But $m$ is the least common multiple and $s$ is less than $m$. So $s=0$.
answered Nov 28 '18 at 11:30
B. Goddard
18.4k21340
Let $m=mbox{lcm}(a,b).$ By division algorithm $r=qm+s$ with $0leq s < m.$ Then $a mid r = qm+s$. Since $a$ also divides $m$, we have $a mid s$. Similarly $bmid s.$ Therefore $s$ is a common multiple of $a$ and $b$. But $m$ is the least common multiple and $s$ is less than $m$. So $s=0$.
answered Nov 28 '18 at 11:30
B. Goddard
18.4k21340
answered Nov 28 '18 at 11:30
B. Goddard
18.4k21340
answered Nov 28 '18 at 11:30
B. Goddard
18.4k21340
answered Nov 28 '18 at 11:30
B. Goddard
18.4k21340
18.4k21340
add a comment |
add a comment |
$text{lcm}(a,b)=q$. By division algorithm, $r=mq+cquad 0leq c<q.$ We have $a|r,a|qimplies a|c$ and $b|r,b|qimplies b|c$. Since $c$ is a common multiple of $a$ and $b$ with $:c<q$. So $c=0$ and $q|r$.
answered Nov 28 '18 at 11:17
Yadati Kiran
1,694619
No. $amid c$ and $bmid c$ does not imply that $abmid c$.
– B. Goddard
Nov 28 '18 at 11:25
@B.Goddard: It was a huge oversight. Thanks.
– Yadati Kiran
Nov 28 '18 at 11:31
add a comment |
$text{lcm}(a,b)=q$. By division algorithm, $r=mq+cquad 0leq c<q.$ We have $a|r,a|qimplies a|c$ and $b|r,b|qimplies b|c$. Since $c$ is a common multiple of $a$ and $b$ with $:c<q$. So $c=0$ and $q|r$.
answered Nov 28 '18 at 11:17
Yadati Kiran
1,694619
No. $amid c$ and $bmid c$ does not imply that $abmid c$.
– B. Goddard
Nov 28 '18 at 11:25
@B.Goddard: It was a huge oversight. Thanks.
– Yadati Kiran
Nov 28 '18 at 11:31
add a comment |
$text{lcm}(a,b)=q$. By division algorithm, $r=mq+cquad 0leq c<q.$ We have $a|r,a|qimplies a|c$ and $b|r,b|qimplies b|c$. Since $c$ is a common multiple of $a$ and $b$ with $:c<q$. So $c=0$ and $q|r$.
answered Nov 28 '18 at 11:17
Yadati Kiran
1,694619
$text{lcm}(a,b)=q$. By division algorithm, $r=mq+cquad 0leq c<q.$ We have $a|r,a|qimplies a|c$ and $b|r,b|qimplies b|c$. Since $c$ is a common multiple of $a$ and $b$ with $:c<q$. So $c=0$ and $q|r$.
answered Nov 28 '18 at 11:17
Yadati Kiran
1,694619
edited Nov 28 '18 at 11:26
answered Nov 28 '18 at 11:17
Yadati Kiran
1,694619
answered Nov 28 '18 at 11:17
Yadati Kiran
1,694619
answered Nov 28 '18 at 11:17
Yadati Kiran
1,694619
1,694619
No. $amid c$ and $bmid c$ does not imply that $abmid c$.
– B. Goddard
Nov 28 '18 at 11:25
@B.Goddard: It was a huge oversight. Thanks.
– Yadati Kiran
Nov 28 '18 at 11:31
add a comment |
No. $amid c$ and $bmid c$ does not imply that $abmid c$.
– B. Goddard
Nov 28 '18 at 11:25
@B.Goddard: It was a huge oversight. Thanks.
– Yadati Kiran
Nov 28 '18 at 11:31
No. $amid c$ and $bmid c$ does not imply that $abmid c$.
– B. Goddard
Nov 28 '18 at 11:25
No. $amid c$ and $bmid c$ does not imply that $abmid c$.
– B. Goddard
Nov 28 '18 at 11:25
@B.Goddard: It was a huge oversight. Thanks.
– Yadati Kiran
Nov 28 '18 at 11:31
@B.Goddard: It was a huge oversight. Thanks.
– Yadati Kiran
Nov 28 '18 at 11:31
add a comment |
Consider the following theorem:
If $m=operatorname{lcm}(a,b)$ then $m|r$ for any common multiple $r$ of $a,b$.
You can, roughly speaking, say that, since $a|r,b|r$, then $r$ is a common multiple of $a,b$. But each common multiple of $a,b$ is a multiple of $operatorname{lcm}(a,b)$. Therefore $r=qoperatorname{lcm}(a,b)$. And the proof completes.
The proof of the theorem can be inferred from @B. Goddard answer.
answered Nov 28 '18 at 13:10
Maged Saeed
8521417
add a comment |
Consider the following theorem:
If $m=operatorname{lcm}(a,b)$ then $m|r$ for any common multiple $r$ of $a,b$.
You can, roughly speaking, say that, since $a|r,b|r$, then $r$ is a common multiple of $a,b$. But each common multiple of $a,b$ is a multiple of $operatorname{lcm}(a,b)$. Therefore $r=qoperatorname{lcm}(a,b)$. And the proof completes.
The proof of the theorem can be inferred from @B. Goddard answer.
answered Nov 28 '18 at 13:10
Maged Saeed
8521417
add a comment |
Consider the following theorem:
If $m=operatorname{lcm}(a,b)$ then $m|r$ for any common multiple $r$ of $a,b$.
You can, roughly speaking, say that, since $a|r,b|r$, then $r$ is a common multiple of $a,b$. But each common multiple of $a,b$ is a multiple of $operatorname{lcm}(a,b)$. Therefore $r=qoperatorname{lcm}(a,b)$. And the proof completes.
The proof of the theorem can be inferred from @B. Goddard answer.
answered Nov 28 '18 at 13:10
Maged Saeed
8521417
Consider the following theorem:
If $m=operatorname{lcm}(a,b)$ then $m|r$ for any common multiple $r$ of $a,b$.
You can, roughly speaking, say that, since $a|r,b|r$, then $r$ is a common multiple of $a,b$. But each common multiple of $a,b$ is a multiple of $operatorname{lcm}(a,b)$. Therefore $r=qoperatorname{lcm}(a,b)$. And the proof completes.
The proof of the theorem can be inferred from @B. Goddard answer.
answered Nov 28 '18 at 13:10
Maged Saeed
8521417
answered Nov 28 '18 at 13:10
Maged Saeed
8521417
answered Nov 28 '18 at 13:10
Maged Saeed
8521417
answered Nov 28 '18 at 13:10
Maged Saeed
8521417
8521417
add a comment |
add a comment |
