Why is $5$ the remainder of $5 over 13$?
$begingroup$
I don't understand why the remainder of $5 over 13$ is $5$. I know that the DA tells us that $5 = 0(13) + r$ so the remainder has to be $5$ based on this, but I'm a little unsure of why/how it works
elementary-number-theory divisibility
edited Dec 9 '18 at 8:25
Batominovski
1
asked Dec 9 '18 at 0:55
mingming
3255
$endgroup$
add a comment |
$begingroup$
I don't understand why the remainder of $5 over 13$ is $5$. I know that the DA tells us that $5 = 0(13) + r$ so the remainder has to be $5$ based on this, but I'm a little unsure of why/how it works
elementary-number-theory divisibility
edited Dec 9 '18 at 8:25
Batominovski
1
asked Dec 9 '18 at 0:55
mingming
3255
$endgroup$
$begingroup$
By definition the remainder of $,5div 13,$ is the least nonnegative element of $, 5+13,Bbb Z$ $ $
$endgroup$
– Bill Dubuque
Dec 9 '18 at 1:06
add a comment |
$begingroup$
I don't understand why the remainder of $5 over 13$ is $5$. I know that the DA tells us that $5 = 0(13) + r$ so the remainder has to be $5$ based on this, but I'm a little unsure of why/how it works
elementary-number-theory divisibility
edited Dec 9 '18 at 8:25
Batominovski
1
asked Dec 9 '18 at 0:55
mingming
3255
$endgroup$
I don't understand why the remainder of $5 over 13$ is $5$. I know that the DA tells us that $5 = 0(13) + r$ so the remainder has to be $5$ based on this, but I'm a little unsure of why/how it works
elementary-number-theory divisibility
elementary-number-theory divisibility
edited Dec 9 '18 at 8:25
Batominovski
1
asked Dec 9 '18 at 0:55
mingming
3255
edited Dec 9 '18 at 8:25
Batominovski
1
asked Dec 9 '18 at 0:55
mingming
3255
edited Dec 9 '18 at 8:25
Batominovski
1
edited Dec 9 '18 at 8:25
Batominovski
1
edited Dec 9 '18 at 8:25
Batominovski
1
1
asked Dec 9 '18 at 0:55
mingming
3255
asked Dec 9 '18 at 0:55
mingming
3255
asked Dec 9 '18 at 0:55
mingming
3255
3255
$begingroup$
By definition the remainder of $,5div 13,$ is the least nonnegative element of $, 5+13,Bbb Z$ $ $
$endgroup$
– Bill Dubuque
Dec 9 '18 at 1:06
add a comment |
$begingroup$
By definition the remainder of $,5div 13,$ is the least nonnegative element of $, 5+13,Bbb Z$ $ $
$endgroup$
– Bill Dubuque
Dec 9 '18 at 1:06
$begingroup$
By definition the remainder of $,5div 13,$ is the least nonnegative element of $, 5+13,Bbb Z$ $ $
$endgroup$
– Bill Dubuque
Dec 9 '18 at 1:06
$begingroup$
By definition the remainder of $,5div 13,$ is the least nonnegative element of $, 5+13,Bbb Z$ $ $
$endgroup$
– Bill Dubuque
Dec 9 '18 at 1:06
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
For positive integers, "divide by $13$" means "subtract the largest multiple of $13$ you can while still remaining nonnegative". The remainder is what's left. If you think in terms of objects, you remove them from the set you start with, in batches of $13$.
If you start with $5$ you can't subtract any multiples of $13$ so all $5$ are left.
You don't need a formal "division algorithm" for this. It works fine if all you know about division is repeated subtraction (which is what the division algorithm formalizes).
answered Dec 9 '18 at 1:00
Ethan BolkerEthan Bolker
42.2k548111
$endgroup$
add a comment |
$begingroup$
For an intuitive approach: If you have five candies to give out equally to 13 children, each child gets 0 candies, and there remain 5 undistributed candies.
answered Dec 9 '18 at 1:04
paw88789paw88789
29.1k12349
$endgroup$
add a comment |
$begingroup$
Well $r$ has to be in a set ${0,1,2...,12}$. Which of this fullfiles equation you wrote?
And this is always true if you divide $a$ by $b$ and $a<b$. You will get $r=a$:
$$ a = 0cdot b +rimplies r=a$$
answered Dec 9 '18 at 0:57
greedoidgreedoid
39k114797
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For positive integers, "divide by $13$" means "subtract the largest multiple of $13$ you can while still remaining nonnegative". The remainder is what's left. If you think in terms of objects, you remove them from the set you start with, in batches of $13$.
If you start with $5$ you can't subtract any multiples of $13$ so all $5$ are left.
You don't need a formal "division algorithm" for this. It works fine if all you know about division is repeated subtraction (which is what the division algorithm formalizes).
answered Dec 9 '18 at 1:00
Ethan BolkerEthan Bolker
42.2k548111
$endgroup$
add a comment |
$begingroup$
For positive integers, "divide by $13$" means "subtract the largest multiple of $13$ you can while still remaining nonnegative". The remainder is what's left. If you think in terms of objects, you remove them from the set you start with, in batches of $13$.
If you start with $5$ you can't subtract any multiples of $13$ so all $5$ are left.
You don't need a formal "division algorithm" for this. It works fine if all you know about division is repeated subtraction (which is what the division algorithm formalizes).
answered Dec 9 '18 at 1:00
Ethan BolkerEthan Bolker
42.2k548111
$endgroup$
add a comment |
$begingroup$
For positive integers, "divide by $13$" means "subtract the largest multiple of $13$ you can while still remaining nonnegative". The remainder is what's left. If you think in terms of objects, you remove them from the set you start with, in batches of $13$.
If you start with $5$ you can't subtract any multiples of $13$ so all $5$ are left.
You don't need a formal "division algorithm" for this. It works fine if all you know about division is repeated subtraction (which is what the division algorithm formalizes).
answered Dec 9 '18 at 1:00
Ethan BolkerEthan Bolker
42.2k548111
$endgroup$
For positive integers, "divide by $13$" means "subtract the largest multiple of $13$ you can while still remaining nonnegative". The remainder is what's left. If you think in terms of objects, you remove them from the set you start with, in batches of $13$.
If you start with $5$ you can't subtract any multiples of $13$ so all $5$ are left.
You don't need a formal "division algorithm" for this. It works fine if all you know about division is repeated subtraction (which is what the division algorithm formalizes).
answered Dec 9 '18 at 1:00
Ethan BolkerEthan Bolker
42.2k548111
answered Dec 9 '18 at 1:00
Ethan BolkerEthan Bolker
42.2k548111
answered Dec 9 '18 at 1:00
Ethan BolkerEthan Bolker
42.2k548111
answered Dec 9 '18 at 1:00
Ethan BolkerEthan Bolker
42.2k548111
42.2k548111
add a comment |
add a comment |
$begingroup$
For an intuitive approach: If you have five candies to give out equally to 13 children, each child gets 0 candies, and there remain 5 undistributed candies.
answered Dec 9 '18 at 1:04
paw88789paw88789
29.1k12349
$endgroup$
add a comment |
$begingroup$
For an intuitive approach: If you have five candies to give out equally to 13 children, each child gets 0 candies, and there remain 5 undistributed candies.
answered Dec 9 '18 at 1:04
paw88789paw88789
29.1k12349
$endgroup$
add a comment |
$begingroup$
For an intuitive approach: If you have five candies to give out equally to 13 children, each child gets 0 candies, and there remain 5 undistributed candies.
answered Dec 9 '18 at 1:04
paw88789paw88789
29.1k12349
$endgroup$
For an intuitive approach: If you have five candies to give out equally to 13 children, each child gets 0 candies, and there remain 5 undistributed candies.
answered Dec 9 '18 at 1:04
paw88789paw88789
29.1k12349
answered Dec 9 '18 at 1:04
paw88789paw88789
29.1k12349
answered Dec 9 '18 at 1:04
paw88789paw88789
29.1k12349
answered Dec 9 '18 at 1:04
paw88789paw88789
29.1k12349
29.1k12349
add a comment |
add a comment |
$begingroup$
Well $r$ has to be in a set ${0,1,2...,12}$. Which of this fullfiles equation you wrote?
And this is always true if you divide $a$ by $b$ and $a<b$. You will get $r=a$:
$$ a = 0cdot b +rimplies r=a$$
answered Dec 9 '18 at 0:57
greedoidgreedoid
39k114797
$endgroup$
add a comment |
$begingroup$
Well $r$ has to be in a set ${0,1,2...,12}$. Which of this fullfiles equation you wrote?
And this is always true if you divide $a$ by $b$ and $a<b$. You will get $r=a$:
$$ a = 0cdot b +rimplies r=a$$
answered Dec 9 '18 at 0:57
greedoidgreedoid
39k114797
$endgroup$
add a comment |
$begingroup$
Well $r$ has to be in a set ${0,1,2...,12}$. Which of this fullfiles equation you wrote?
And this is always true if you divide $a$ by $b$ and $a<b$. You will get $r=a$:
$$ a = 0cdot b +rimplies r=a$$
answered Dec 9 '18 at 0:57
greedoidgreedoid
39k114797
$endgroup$
Well $r$ has to be in a set ${0,1,2...,12}$. Which of this fullfiles equation you wrote?
And this is always true if you divide $a$ by $b$ and $a<b$. You will get $r=a$:
$$ a = 0cdot b +rimplies r=a$$
answered Dec 9 '18 at 0:57
greedoidgreedoid
39k114797
answered Dec 9 '18 at 0:57
greedoidgreedoid
39k114797
answered Dec 9 '18 at 0:57
greedoidgreedoid
39k114797
answered Dec 9 '18 at 0:57
greedoidgreedoid
39k114797
39k114797
add a comment |
add a comment |
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$begingroup$
By definition the remainder of $,5div 13,$ is the least nonnegative element of $, 5+13,Bbb Z$ $ $
$endgroup$
– Bill Dubuque
Dec 9 '18 at 1:06