Proof of modulo property
There is a video on youtube where a guy in the comments proves the following:
If $15l equiv 2 mod7$, then $l equiv 2 mod7$.
He does it like this:
15L = 2 (mod 7)
=> 15L = 7k + 2 for some k in the integers
Let k = 2T where T is an integer
=> 15L = 14T + 2
=> L = 14T - 14L + 2
=> L = 7(2T - 2L) + 2
Let H = (2T - 2L), then H is an integer.
=> L = 7H + 2
=> L = 2 (mod 7)
What bothers me is the following line "Let k = 2T where T is an integer". Why replace $k$ with a multiple of 2? We would get the same result if we do not even replace $k$ and leave it as it is for example:
L = 7k - 14L + 2
=> L = 7(k - 2L) + 2
=> L = 2 (mod 7)
Is my method correct as well or is there some deeper reasoning as to why would he replace $k$ with $2T$?
EDIT:
Here is the video if anyone is interested, the comment is made by the user RB:
https://www.youtube.com/watch?v=LInNgWMtFEs&lc=z23ts1qyloeagpzhm04t1aokgn15f4y4gqsns5m1d5p3rk0h00410.1543158282809495
modular-arithmetic proof-explanation
add a comment |
There is a video on youtube where a guy in the comments proves the following:
If $15l equiv 2 mod7$, then $l equiv 2 mod7$.
He does it like this:
15L = 2 (mod 7)
=> 15L = 7k + 2 for some k in the integers
Let k = 2T where T is an integer
=> 15L = 14T + 2
=> L = 14T - 14L + 2
=> L = 7(2T - 2L) + 2
Let H = (2T - 2L), then H is an integer.
=> L = 7H + 2
=> L = 2 (mod 7)
What bothers me is the following line "Let k = 2T where T is an integer". Why replace $k$ with a multiple of 2? We would get the same result if we do not even replace $k$ and leave it as it is for example:
L = 7k - 14L + 2
=> L = 7(k - 2L) + 2
=> L = 2 (mod 7)
Is my method correct as well or is there some deeper reasoning as to why would he replace $k$ with $2T$?
EDIT:
Here is the video if anyone is interested, the comment is made by the user RB:
https://www.youtube.com/watch?v=LInNgWMtFEs&lc=z23ts1qyloeagpzhm04t1aokgn15f4y4gqsns5m1d5p3rk0h00410.1543158282809495
modular-arithmetic proof-explanation
add a comment |
There is a video on youtube where a guy in the comments proves the following:
If $15l equiv 2 mod7$, then $l equiv 2 mod7$.
He does it like this:
15L = 2 (mod 7)
=> 15L = 7k + 2 for some k in the integers
Let k = 2T where T is an integer
=> 15L = 14T + 2
=> L = 14T - 14L + 2
=> L = 7(2T - 2L) + 2
Let H = (2T - 2L), then H is an integer.
=> L = 7H + 2
=> L = 2 (mod 7)
What bothers me is the following line "Let k = 2T where T is an integer". Why replace $k$ with a multiple of 2? We would get the same result if we do not even replace $k$ and leave it as it is for example:
L = 7k - 14L + 2
=> L = 7(k - 2L) + 2
=> L = 2 (mod 7)
Is my method correct as well or is there some deeper reasoning as to why would he replace $k$ with $2T$?
EDIT:
Here is the video if anyone is interested, the comment is made by the user RB:
https://www.youtube.com/watch?v=LInNgWMtFEs&lc=z23ts1qyloeagpzhm04t1aokgn15f4y4gqsns5m1d5p3rk0h00410.1543158282809495
modular-arithmetic proof-explanation
There is a video on youtube where a guy in the comments proves the following:
If $15l equiv 2 mod7$, then $l equiv 2 mod7$.
He does it like this:
15L = 2 (mod 7)
=> 15L = 7k + 2 for some k in the integers
Let k = 2T where T is an integer
=> 15L = 14T + 2
=> L = 14T - 14L + 2
=> L = 7(2T - 2L) + 2
Let H = (2T - 2L), then H is an integer.
=> L = 7H + 2
=> L = 2 (mod 7)
What bothers me is the following line "Let k = 2T where T is an integer". Why replace $k$ with a multiple of 2? We would get the same result if we do not even replace $k$ and leave it as it is for example:
L = 7k - 14L + 2
=> L = 7(k - 2L) + 2
=> L = 2 (mod 7)
Is my method correct as well or is there some deeper reasoning as to why would he replace $k$ with $2T$?
EDIT:
Here is the video if anyone is interested, the comment is made by the user RB:
https://www.youtube.com/watch?v=LInNgWMtFEs&lc=z23ts1qyloeagpzhm04t1aokgn15f4y4gqsns5m1d5p3rk0h00410.1543158282809495
modular-arithmetic proof-explanation
modular-arithmetic proof-explanation
edited Dec 7 '18 at 18:09
asked Dec 7 '18 at 16:06
Michael Munta
568
568
add a comment |
add a comment |
5 Answers
5
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I'd say you are right and the video is wrong. For instance, we could have $k=19$, which can't be written as $k=2T$ for an integer $T$.
By the way, a possibly simpler approach to the whole thing is to note that $15equiv 1 pmod{7}$.
So $2equiv 15lequiv l pmod{7}$.
add a comment |
Your concerns about the video are justified.
E.g. we have $15times 9=135=7times 19+2$ but there is no integer $k$ such that $15times 9=135=14k+2$.
Your method is okay.
On base of $7mid 14l$ you can also observe that: $$7mid 15l-2iff7mid 14l+l-2iff 7mid l-2$$
I am not sure what does your last sentence prove. Can you elaborate?
– Michael Munta
Dec 9 '18 at 8:22
I think the last part should be $7 | l - 2$
– Michael Munta
Dec 9 '18 at 8:36
@MichaelMunta Indeed that was a typo. Repaired now. Thank you for attending me. Is everything clear to you now?
– drhab
Dec 9 '18 at 11:17
No problem. I am just not sure how $14l$ disappears from $7|14l + l - 2$. Can you explain?
– Michael Munta
Dec 9 '18 at 11:34
1
$7mid a$ together with $7mid b$ implies that $7|a-b$. That can be applied on $a=14l+l-2$ and $b=14l$.
– drhab
Dec 9 '18 at 13:13
add a comment |
The argument is incorrect since $, 15,l = 7,k+2,$ does not imply $,2mid k,,$ (e.g. $ l,k = -5,-11$). Further, the argument uses unidirectional inferences where bidirectional inferences are required. Below is one correct way to do the proof in that manner.
$$begin{align}
15, l &equiv 2!pmod{! 7}\
iff exists, k!: 15,l &= 2+7,k\
iff exists, k!: l &= 2+7(k!-!2l)\
iff exists, j!: l &= 2+7,j\
iffqquadquad , l &equiv 2!pmod{! 7}
end{align}qquadqquad$$
It's simpler to use basic rules of modular arithmetic. By the Congruence Product Rule we deduce
$!bmod 7!:, color{#c00}{15equiv 1},Rightarrow, color{#c00}{15},lequiv color{#c00}1,lequiv l $ thus $ 2equiv 15,lequiv l$
add a comment |
If someone in their comments indeed says that, then your doubts are well justified — that step is plain wrong. In the setting of this question, $k$ does NOT have to be an even number, so we can NOT (in general) set it to be $2t$ for an integer $t$.
Quick example: if $l=9$, then $15l=15cdot9=135equiv2 mod7$, but then $k=133/7=19$ can't be represented as "$k=2t$ for an integer $t$".
Your solution, however, is perfectly correct!
add a comment |
Why not using
$color{blue}{15 equiv 1 mod 7}$?
$$Rightarrow color{blue}{15}l equiv color{blue}{1}l equiv 2 mod 7$$
add a comment |
Your Answer
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
I'd say you are right and the video is wrong. For instance, we could have $k=19$, which can't be written as $k=2T$ for an integer $T$.
By the way, a possibly simpler approach to the whole thing is to note that $15equiv 1 pmod{7}$.
So $2equiv 15lequiv l pmod{7}$.
add a comment |
I'd say you are right and the video is wrong. For instance, we could have $k=19$, which can't be written as $k=2T$ for an integer $T$.
By the way, a possibly simpler approach to the whole thing is to note that $15equiv 1 pmod{7}$.
So $2equiv 15lequiv l pmod{7}$.
add a comment |
I'd say you are right and the video is wrong. For instance, we could have $k=19$, which can't be written as $k=2T$ for an integer $T$.
By the way, a possibly simpler approach to the whole thing is to note that $15equiv 1 pmod{7}$.
So $2equiv 15lequiv l pmod{7}$.
I'd say you are right and the video is wrong. For instance, we could have $k=19$, which can't be written as $k=2T$ for an integer $T$.
By the way, a possibly simpler approach to the whole thing is to note that $15equiv 1 pmod{7}$.
So $2equiv 15lequiv l pmod{7}$.
answered Dec 7 '18 at 16:13
paw88789
29k12349
29k12349
add a comment |
add a comment |
Your concerns about the video are justified.
E.g. we have $15times 9=135=7times 19+2$ but there is no integer $k$ such that $15times 9=135=14k+2$.
Your method is okay.
On base of $7mid 14l$ you can also observe that: $$7mid 15l-2iff7mid 14l+l-2iff 7mid l-2$$
I am not sure what does your last sentence prove. Can you elaborate?
– Michael Munta
Dec 9 '18 at 8:22
I think the last part should be $7 | l - 2$
– Michael Munta
Dec 9 '18 at 8:36
@MichaelMunta Indeed that was a typo. Repaired now. Thank you for attending me. Is everything clear to you now?
– drhab
Dec 9 '18 at 11:17
No problem. I am just not sure how $14l$ disappears from $7|14l + l - 2$. Can you explain?
– Michael Munta
Dec 9 '18 at 11:34
1
$7mid a$ together with $7mid b$ implies that $7|a-b$. That can be applied on $a=14l+l-2$ and $b=14l$.
– drhab
Dec 9 '18 at 13:13
add a comment |
Your concerns about the video are justified.
E.g. we have $15times 9=135=7times 19+2$ but there is no integer $k$ such that $15times 9=135=14k+2$.
Your method is okay.
On base of $7mid 14l$ you can also observe that: $$7mid 15l-2iff7mid 14l+l-2iff 7mid l-2$$
I am not sure what does your last sentence prove. Can you elaborate?
– Michael Munta
Dec 9 '18 at 8:22
I think the last part should be $7 | l - 2$
– Michael Munta
Dec 9 '18 at 8:36
@MichaelMunta Indeed that was a typo. Repaired now. Thank you for attending me. Is everything clear to you now?
– drhab
Dec 9 '18 at 11:17
No problem. I am just not sure how $14l$ disappears from $7|14l + l - 2$. Can you explain?
– Michael Munta
Dec 9 '18 at 11:34
1
$7mid a$ together with $7mid b$ implies that $7|a-b$. That can be applied on $a=14l+l-2$ and $b=14l$.
– drhab
Dec 9 '18 at 13:13
add a comment |
Your concerns about the video are justified.
E.g. we have $15times 9=135=7times 19+2$ but there is no integer $k$ such that $15times 9=135=14k+2$.
Your method is okay.
On base of $7mid 14l$ you can also observe that: $$7mid 15l-2iff7mid 14l+l-2iff 7mid l-2$$
Your concerns about the video are justified.
E.g. we have $15times 9=135=7times 19+2$ but there is no integer $k$ such that $15times 9=135=14k+2$.
Your method is okay.
On base of $7mid 14l$ you can also observe that: $$7mid 15l-2iff7mid 14l+l-2iff 7mid l-2$$
edited Dec 9 '18 at 11:15
answered Dec 7 '18 at 16:20
drhab
97.9k544129
97.9k544129
I am not sure what does your last sentence prove. Can you elaborate?
– Michael Munta
Dec 9 '18 at 8:22
I think the last part should be $7 | l - 2$
– Michael Munta
Dec 9 '18 at 8:36
@MichaelMunta Indeed that was a typo. Repaired now. Thank you for attending me. Is everything clear to you now?
– drhab
Dec 9 '18 at 11:17
No problem. I am just not sure how $14l$ disappears from $7|14l + l - 2$. Can you explain?
– Michael Munta
Dec 9 '18 at 11:34
1
$7mid a$ together with $7mid b$ implies that $7|a-b$. That can be applied on $a=14l+l-2$ and $b=14l$.
– drhab
Dec 9 '18 at 13:13
add a comment |
I am not sure what does your last sentence prove. Can you elaborate?
– Michael Munta
Dec 9 '18 at 8:22
I think the last part should be $7 | l - 2$
– Michael Munta
Dec 9 '18 at 8:36
@MichaelMunta Indeed that was a typo. Repaired now. Thank you for attending me. Is everything clear to you now?
– drhab
Dec 9 '18 at 11:17
No problem. I am just not sure how $14l$ disappears from $7|14l + l - 2$. Can you explain?
– Michael Munta
Dec 9 '18 at 11:34
1
$7mid a$ together with $7mid b$ implies that $7|a-b$. That can be applied on $a=14l+l-2$ and $b=14l$.
– drhab
Dec 9 '18 at 13:13
I am not sure what does your last sentence prove. Can you elaborate?
– Michael Munta
Dec 9 '18 at 8:22
I am not sure what does your last sentence prove. Can you elaborate?
– Michael Munta
Dec 9 '18 at 8:22
I think the last part should be $7 | l - 2$
– Michael Munta
Dec 9 '18 at 8:36
I think the last part should be $7 | l - 2$
– Michael Munta
Dec 9 '18 at 8:36
@MichaelMunta Indeed that was a typo. Repaired now. Thank you for attending me. Is everything clear to you now?
– drhab
Dec 9 '18 at 11:17
@MichaelMunta Indeed that was a typo. Repaired now. Thank you for attending me. Is everything clear to you now?
– drhab
Dec 9 '18 at 11:17
No problem. I am just not sure how $14l$ disappears from $7|14l + l - 2$. Can you explain?
– Michael Munta
Dec 9 '18 at 11:34
No problem. I am just not sure how $14l$ disappears from $7|14l + l - 2$. Can you explain?
– Michael Munta
Dec 9 '18 at 11:34
1
1
$7mid a$ together with $7mid b$ implies that $7|a-b$. That can be applied on $a=14l+l-2$ and $b=14l$.
– drhab
Dec 9 '18 at 13:13
$7mid a$ together with $7mid b$ implies that $7|a-b$. That can be applied on $a=14l+l-2$ and $b=14l$.
– drhab
Dec 9 '18 at 13:13
add a comment |
The argument is incorrect since $, 15,l = 7,k+2,$ does not imply $,2mid k,,$ (e.g. $ l,k = -5,-11$). Further, the argument uses unidirectional inferences where bidirectional inferences are required. Below is one correct way to do the proof in that manner.
$$begin{align}
15, l &equiv 2!pmod{! 7}\
iff exists, k!: 15,l &= 2+7,k\
iff exists, k!: l &= 2+7(k!-!2l)\
iff exists, j!: l &= 2+7,j\
iffqquadquad , l &equiv 2!pmod{! 7}
end{align}qquadqquad$$
It's simpler to use basic rules of modular arithmetic. By the Congruence Product Rule we deduce
$!bmod 7!:, color{#c00}{15equiv 1},Rightarrow, color{#c00}{15},lequiv color{#c00}1,lequiv l $ thus $ 2equiv 15,lequiv l$
add a comment |
The argument is incorrect since $, 15,l = 7,k+2,$ does not imply $,2mid k,,$ (e.g. $ l,k = -5,-11$). Further, the argument uses unidirectional inferences where bidirectional inferences are required. Below is one correct way to do the proof in that manner.
$$begin{align}
15, l &equiv 2!pmod{! 7}\
iff exists, k!: 15,l &= 2+7,k\
iff exists, k!: l &= 2+7(k!-!2l)\
iff exists, j!: l &= 2+7,j\
iffqquadquad , l &equiv 2!pmod{! 7}
end{align}qquadqquad$$
It's simpler to use basic rules of modular arithmetic. By the Congruence Product Rule we deduce
$!bmod 7!:, color{#c00}{15equiv 1},Rightarrow, color{#c00}{15},lequiv color{#c00}1,lequiv l $ thus $ 2equiv 15,lequiv l$
add a comment |
The argument is incorrect since $, 15,l = 7,k+2,$ does not imply $,2mid k,,$ (e.g. $ l,k = -5,-11$). Further, the argument uses unidirectional inferences where bidirectional inferences are required. Below is one correct way to do the proof in that manner.
$$begin{align}
15, l &equiv 2!pmod{! 7}\
iff exists, k!: 15,l &= 2+7,k\
iff exists, k!: l &= 2+7(k!-!2l)\
iff exists, j!: l &= 2+7,j\
iffqquadquad , l &equiv 2!pmod{! 7}
end{align}qquadqquad$$
It's simpler to use basic rules of modular arithmetic. By the Congruence Product Rule we deduce
$!bmod 7!:, color{#c00}{15equiv 1},Rightarrow, color{#c00}{15},lequiv color{#c00}1,lequiv l $ thus $ 2equiv 15,lequiv l$
The argument is incorrect since $, 15,l = 7,k+2,$ does not imply $,2mid k,,$ (e.g. $ l,k = -5,-11$). Further, the argument uses unidirectional inferences where bidirectional inferences are required. Below is one correct way to do the proof in that manner.
$$begin{align}
15, l &equiv 2!pmod{! 7}\
iff exists, k!: 15,l &= 2+7,k\
iff exists, k!: l &= 2+7(k!-!2l)\
iff exists, j!: l &= 2+7,j\
iffqquadquad , l &equiv 2!pmod{! 7}
end{align}qquadqquad$$
It's simpler to use basic rules of modular arithmetic. By the Congruence Product Rule we deduce
$!bmod 7!:, color{#c00}{15equiv 1},Rightarrow, color{#c00}{15},lequiv color{#c00}1,lequiv l $ thus $ 2equiv 15,lequiv l$
answered Dec 7 '18 at 20:26
Bill Dubuque
208k29190628
208k29190628
add a comment |
add a comment |
If someone in their comments indeed says that, then your doubts are well justified — that step is plain wrong. In the setting of this question, $k$ does NOT have to be an even number, so we can NOT (in general) set it to be $2t$ for an integer $t$.
Quick example: if $l=9$, then $15l=15cdot9=135equiv2 mod7$, but then $k=133/7=19$ can't be represented as "$k=2t$ for an integer $t$".
Your solution, however, is perfectly correct!
add a comment |
If someone in their comments indeed says that, then your doubts are well justified — that step is plain wrong. In the setting of this question, $k$ does NOT have to be an even number, so we can NOT (in general) set it to be $2t$ for an integer $t$.
Quick example: if $l=9$, then $15l=15cdot9=135equiv2 mod7$, but then $k=133/7=19$ can't be represented as "$k=2t$ for an integer $t$".
Your solution, however, is perfectly correct!
add a comment |
If someone in their comments indeed says that, then your doubts are well justified — that step is plain wrong. In the setting of this question, $k$ does NOT have to be an even number, so we can NOT (in general) set it to be $2t$ for an integer $t$.
Quick example: if $l=9$, then $15l=15cdot9=135equiv2 mod7$, but then $k=133/7=19$ can't be represented as "$k=2t$ for an integer $t$".
Your solution, however, is perfectly correct!
If someone in their comments indeed says that, then your doubts are well justified — that step is plain wrong. In the setting of this question, $k$ does NOT have to be an even number, so we can NOT (in general) set it to be $2t$ for an integer $t$.
Quick example: if $l=9$, then $15l=15cdot9=135equiv2 mod7$, but then $k=133/7=19$ can't be represented as "$k=2t$ for an integer $t$".
Your solution, however, is perfectly correct!
edited Dec 8 '18 at 17:17
answered Dec 7 '18 at 16:15
zipirovich
11k11631
11k11631
add a comment |
add a comment |
Why not using
$color{blue}{15 equiv 1 mod 7}$?
$$Rightarrow color{blue}{15}l equiv color{blue}{1}l equiv 2 mod 7$$
add a comment |
Why not using
$color{blue}{15 equiv 1 mod 7}$?
$$Rightarrow color{blue}{15}l equiv color{blue}{1}l equiv 2 mod 7$$
add a comment |
Why not using
$color{blue}{15 equiv 1 mod 7}$?
$$Rightarrow color{blue}{15}l equiv color{blue}{1}l equiv 2 mod 7$$
Why not using
$color{blue}{15 equiv 1 mod 7}$?
$$Rightarrow color{blue}{15}l equiv color{blue}{1}l equiv 2 mod 7$$
answered Dec 7 '18 at 16:13
trancelocation
9,1051521
9,1051521
add a comment |
add a comment |
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