Prove $forall a in mathbb{N}, forall b in mathbb{N}, ab text{ mod } n = (a text{ mod } n)(b text{ mod } n)...
This question already has an answer here:
Product Rule for the mod operator
2 answers
Please prove $forall a in mathbb{N}, forall b in mathbb{N}, ab text{ mod } n = (a text{ mod } n)(b text{ mod } n) text{ mod } n$. I have been stuck on this problem for quite a while now. I have no idea how to approach this problem. Alongside proving this, can someone please give me advice as to how I should tackle these types of proofs.
elementary-number-theory discrete-mathematics
marked as duplicate by Bill Dubuque, Lord Shark the Unknown, Rebellos, Jyrki Lahtonen, Lee David Chung Lin Dec 4 '18 at 21:16
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.
|
show 1 more comment
This question already has an answer here:
Product Rule for the mod operator
2 answers
Please prove $forall a in mathbb{N}, forall b in mathbb{N}, ab text{ mod } n = (a text{ mod } n)(b text{ mod } n) text{ mod } n$. I have been stuck on this problem for quite a while now. I have no idea how to approach this problem. Alongside proving this, can someone please give me advice as to how I should tackle these types of proofs.
elementary-number-theory discrete-mathematics
marked as duplicate by Bill Dubuque, Lord Shark the Unknown, Rebellos, Jyrki Lahtonen, Lee David Chung Lin Dec 4 '18 at 21:16
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.
1
$(amod n)(bmod n)mod n$ has no sense...
– Surb
Dec 4 '18 at 16:08
1
@Surb Why not? $amod n$ and $bmod n$ are natural numbers.
– mfl
Dec 4 '18 at 16:12
1
@mfl: Let denote $[a]$ for $(amod n)$. Then $[a]$ is an equivalence class, not a natural number. If for you $[a]mod n$ make sense, you should tel me which one... in one side you have an element of $mathbb Z/_sim$ and on the other hand an relation on $mathbb Z$...
– Surb
Dec 4 '18 at 16:21
@Surb I agree. But if $a,ninmathbb N$ you can identify $amod n$ with the smallest element of the equivalence class and thus you get a natural number.
– mfl
Dec 4 '18 at 16:22
@mfl: indeed we understand... but why not writing it correctly from the beginning instead of writing something wrong and suppose what it could be ?
– Surb
Dec 4 '18 at 16:32
|
show 1 more comment
This question already has an answer here:
Product Rule for the mod operator
2 answers
Please prove $forall a in mathbb{N}, forall b in mathbb{N}, ab text{ mod } n = (a text{ mod } n)(b text{ mod } n) text{ mod } n$. I have been stuck on this problem for quite a while now. I have no idea how to approach this problem. Alongside proving this, can someone please give me advice as to how I should tackle these types of proofs.
elementary-number-theory discrete-mathematics
This question already has an answer here:
Product Rule for the mod operator
2 answers
Please prove $forall a in mathbb{N}, forall b in mathbb{N}, ab text{ mod } n = (a text{ mod } n)(b text{ mod } n) text{ mod } n$. I have been stuck on this problem for quite a while now. I have no idea how to approach this problem. Alongside proving this, can someone please give me advice as to how I should tackle these types of proofs.
This question already has an answer here:
Product Rule for the mod operator
2 answers
elementary-number-theory discrete-mathematics
elementary-number-theory discrete-mathematics
edited Dec 4 '18 at 18:18
Bill Dubuque
209k29190629
209k29190629
asked Dec 4 '18 at 16:01
Noobcoder
274
274
marked as duplicate by Bill Dubuque, Lord Shark the Unknown, Rebellos, Jyrki Lahtonen, Lee David Chung Lin Dec 4 '18 at 21:16
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, Lord Shark the Unknown, Rebellos, Jyrki Lahtonen, Lee David Chung Lin Dec 4 '18 at 21:16
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.
1
$(amod n)(bmod n)mod n$ has no sense...
– Surb
Dec 4 '18 at 16:08
1
@Surb Why not? $amod n$ and $bmod n$ are natural numbers.
– mfl
Dec 4 '18 at 16:12
1
@mfl: Let denote $[a]$ for $(amod n)$. Then $[a]$ is an equivalence class, not a natural number. If for you $[a]mod n$ make sense, you should tel me which one... in one side you have an element of $mathbb Z/_sim$ and on the other hand an relation on $mathbb Z$...
– Surb
Dec 4 '18 at 16:21
@Surb I agree. But if $a,ninmathbb N$ you can identify $amod n$ with the smallest element of the equivalence class and thus you get a natural number.
– mfl
Dec 4 '18 at 16:22
@mfl: indeed we understand... but why not writing it correctly from the beginning instead of writing something wrong and suppose what it could be ?
– Surb
Dec 4 '18 at 16:32
|
show 1 more comment
1
$(amod n)(bmod n)mod n$ has no sense...
– Surb
Dec 4 '18 at 16:08
1
@Surb Why not? $amod n$ and $bmod n$ are natural numbers.
– mfl
Dec 4 '18 at 16:12
1
@mfl: Let denote $[a]$ for $(amod n)$. Then $[a]$ is an equivalence class, not a natural number. If for you $[a]mod n$ make sense, you should tel me which one... in one side you have an element of $mathbb Z/_sim$ and on the other hand an relation on $mathbb Z$...
– Surb
Dec 4 '18 at 16:21
@Surb I agree. But if $a,ninmathbb N$ you can identify $amod n$ with the smallest element of the equivalence class and thus you get a natural number.
– mfl
Dec 4 '18 at 16:22
@mfl: indeed we understand... but why not writing it correctly from the beginning instead of writing something wrong and suppose what it could be ?
– Surb
Dec 4 '18 at 16:32
1
1
$(amod n)(bmod n)mod n$ has no sense...
– Surb
Dec 4 '18 at 16:08
$(amod n)(bmod n)mod n$ has no sense...
– Surb
Dec 4 '18 at 16:08
1
1
@Surb Why not? $amod n$ and $bmod n$ are natural numbers.
– mfl
Dec 4 '18 at 16:12
@Surb Why not? $amod n$ and $bmod n$ are natural numbers.
– mfl
Dec 4 '18 at 16:12
1
1
@mfl: Let denote $[a]$ for $(amod n)$. Then $[a]$ is an equivalence class, not a natural number. If for you $[a]mod n$ make sense, you should tel me which one... in one side you have an element of $mathbb Z/_sim$ and on the other hand an relation on $mathbb Z$...
– Surb
Dec 4 '18 at 16:21
@mfl: Let denote $[a]$ for $(amod n)$. Then $[a]$ is an equivalence class, not a natural number. If for you $[a]mod n$ make sense, you should tel me which one... in one side you have an element of $mathbb Z/_sim$ and on the other hand an relation on $mathbb Z$...
– Surb
Dec 4 '18 at 16:21
@Surb I agree. But if $a,ninmathbb N$ you can identify $amod n$ with the smallest element of the equivalence class and thus you get a natural number.
– mfl
Dec 4 '18 at 16:22
@Surb I agree. But if $a,ninmathbb N$ you can identify $amod n$ with the smallest element of the equivalence class and thus you get a natural number.
– mfl
Dec 4 '18 at 16:22
@mfl: indeed we understand... but why not writing it correctly from the beginning instead of writing something wrong and suppose what it could be ?
– Surb
Dec 4 '18 at 16:32
@mfl: indeed we understand... but why not writing it correctly from the beginning instead of writing something wrong and suppose what it could be ?
– Surb
Dec 4 '18 at 16:32
|
show 1 more comment
2 Answers
2
active
oldest
votes
Let $x=a (text{mod} n), y=b (text{mod} n).$ Then for some integers $k,h$ the numbers $a,b$ decompose as $a=ncdot k+x, b=ncdot l+y.$
So, for this reason $$ab=(nk+x)(nl+y)=xy+nlx+nky+n^2lk=xy+n(lx+ky+nlk).$$
Therefore, $ab (text{mod} n)=xy=big(a (text{mod} n)cdot b (text{mod} n)big) (text{mod} n).$
add a comment |
Hint
Given $a,b,ninmathbb N$ there exist $c,d,r,sinmathbb N$ such that $a=c+rn, b=d+sn.$ And it is
$$ab=cd+(cs+rd+rsn)n.$$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $x=a (text{mod} n), y=b (text{mod} n).$ Then for some integers $k,h$ the numbers $a,b$ decompose as $a=ncdot k+x, b=ncdot l+y.$
So, for this reason $$ab=(nk+x)(nl+y)=xy+nlx+nky+n^2lk=xy+n(lx+ky+nlk).$$
Therefore, $ab (text{mod} n)=xy=big(a (text{mod} n)cdot b (text{mod} n)big) (text{mod} n).$
add a comment |
Let $x=a (text{mod} n), y=b (text{mod} n).$ Then for some integers $k,h$ the numbers $a,b$ decompose as $a=ncdot k+x, b=ncdot l+y.$
So, for this reason $$ab=(nk+x)(nl+y)=xy+nlx+nky+n^2lk=xy+n(lx+ky+nlk).$$
Therefore, $ab (text{mod} n)=xy=big(a (text{mod} n)cdot b (text{mod} n)big) (text{mod} n).$
add a comment |
Let $x=a (text{mod} n), y=b (text{mod} n).$ Then for some integers $k,h$ the numbers $a,b$ decompose as $a=ncdot k+x, b=ncdot l+y.$
So, for this reason $$ab=(nk+x)(nl+y)=xy+nlx+nky+n^2lk=xy+n(lx+ky+nlk).$$
Therefore, $ab (text{mod} n)=xy=big(a (text{mod} n)cdot b (text{mod} n)big) (text{mod} n).$
Let $x=a (text{mod} n), y=b (text{mod} n).$ Then for some integers $k,h$ the numbers $a,b$ decompose as $a=ncdot k+x, b=ncdot l+y.$
So, for this reason $$ab=(nk+x)(nl+y)=xy+nlx+nky+n^2lk=xy+n(lx+ky+nlk).$$
Therefore, $ab (text{mod} n)=xy=big(a (text{mod} n)cdot b (text{mod} n)big) (text{mod} n).$
edited Dec 4 '18 at 16:32
answered Dec 4 '18 at 16:09
Riccardo Ceccon
1,040321
1,040321
add a comment |
add a comment |
Hint
Given $a,b,ninmathbb N$ there exist $c,d,r,sinmathbb N$ such that $a=c+rn, b=d+sn.$ And it is
$$ab=cd+(cs+rd+rsn)n.$$
add a comment |
Hint
Given $a,b,ninmathbb N$ there exist $c,d,r,sinmathbb N$ such that $a=c+rn, b=d+sn.$ And it is
$$ab=cd+(cs+rd+rsn)n.$$
add a comment |
Hint
Given $a,b,ninmathbb N$ there exist $c,d,r,sinmathbb N$ such that $a=c+rn, b=d+sn.$ And it is
$$ab=cd+(cs+rd+rsn)n.$$
Hint
Given $a,b,ninmathbb N$ there exist $c,d,r,sinmathbb N$ such that $a=c+rn, b=d+sn.$ And it is
$$ab=cd+(cs+rd+rsn)n.$$
answered Dec 4 '18 at 16:08
mfl
26k12141
26k12141
add a comment |
add a comment |
1
$(amod n)(bmod n)mod n$ has no sense...
– Surb
Dec 4 '18 at 16:08
1
@Surb Why not? $amod n$ and $bmod n$ are natural numbers.
– mfl
Dec 4 '18 at 16:12
1
@mfl: Let denote $[a]$ for $(amod n)$. Then $[a]$ is an equivalence class, not a natural number. If for you $[a]mod n$ make sense, you should tel me which one... in one side you have an element of $mathbb Z/_sim$ and on the other hand an relation on $mathbb Z$...
– Surb
Dec 4 '18 at 16:21
@Surb I agree. But if $a,ninmathbb N$ you can identify $amod n$ with the smallest element of the equivalence class and thus you get a natural number.
– mfl
Dec 4 '18 at 16:22
@mfl: indeed we understand... but why not writing it correctly from the beginning instead of writing something wrong and suppose what it could be ?
– Surb
Dec 4 '18 at 16:32