Proof of modulo property

Question

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

score 4 · Accepted Answer · answered Dec 07 '18 at 16:13

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 $15\equiv 1 \pmod{7}$.

So $2\equiv 15l\equiv l \pmod{7}$.

drhab · Answer 2 · 2018-12-09T11:15:23.070

4

Your concerns about the video are justified.

E.g. we have $15\times 9=135=7\times 19+2$ but there is no integer $k$ such that $15\times 9=135=14k+2$.

Your method is okay.

On base of $7\mid 14l$ you can also observe that: $$7\mid 15l-2\iff7\mid 14l+l-2\iff 7\mid l-2$$

edited Dec 09 '18 at 11:15

answered Dec 07 '18 at 16:20

drhab

151,093

I am not sure what does your last sentence prove. Can you elaborate? – Michael Munta Dec 09 '18 at 08:22
I think the last part should be $7 | l - 2$ – Michael Munta Dec 09 '18 at 08:36
@MichaelMunta Indeed that was a typo. Repaired now. Thank you for attending me. Is everything clear to you now? – drhab Dec 09 '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 09 '18 at 11:34
1

$7\mid a$ together with $7\mid b$ implies that $7|a-b$. That can be applied on $a=14l+l-2$ and $b=14l$. – drhab Dec 09 '18 at 13:13

score 3 · Answer 3 · answered Dec 07 '18 at 20:26

The argument is incorrect since $\, 15\,l = 7\,k+2\,$ does not imply $\,2\mid k\,\,$ (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.

$$\begin{align} 15\, l &\equiv 2\!\pmod{\! 7}\\ \iff \exists\, k\!:\ 15\,l &= 2+7\,k\\ \iff \exists\, k\!:\ \ \ \ \ \ l &= 2+7(k\!-\!2l)\\ \iff \exists\, j\!:\ \ \ \ \ \ l &= 2+7\,j\\ \iff\qquad\quad\ \, l &\equiv 2\!\pmod{\! 7} \end{align}\qquad\qquad$$

It's simpler to use basic rules of modular arithmetic. By the Congruence Product Rule we deduce

$\!\bmod 7\!:\,\ \color{#c00}{15\equiv 1}\,\Rightarrow\, \color{#c00}{15}\,l\equiv \color{#c00}1\,l\equiv l\ $ thus $\ 2\equiv 15\,l\equiv l$

zipirovich · Answer 4 · 2018-12-08T17:17:03.840

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 $15l=15\cdot9=135\equiv2 \mod7$, but then $k=133/7=19$ can't be represented as "$k=2t$ for an integer $t$".

Your solution, however, is perfectly correct!

score 0 · Answer 5 · answered Dec 07 '18 at 16:13

0

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 07 '18 at 16:13

trancelocation

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Proof of modulo property

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