writing a proof

[kory]

I'm trying to prove that For all integers n, m, and p, if n| (72m–p) and n| (8m), then n| p
but Ive been stuck for hours trying to figure out what to do.

 

[JeffM]

[MATH]\text {Given } n\ | \ (72m - p) \text { and } n \ | \ 8m.[/MATH]
USE YOUR DEFINITIONS.

[MATH]\text {By definition, } n \ | \ 8m \iff \exists \text { integer } u \text { such that } u = \dfrac{8m}{n}.[/MATH]
[MATH]\text {By definition, } n \ | \ (9m - p) \iff \exists \text { integer } v \text { such that } v = \dfrac{72m - p}{n}.[/MATH]
[MATH]\therefore v = \dfrac{72m}{n} - \dfrac{p}{n} \implies[/MATH]
[MATH]\dfrac{72m}{n} - v = \dfrac{p}{n} \implies[/MATH]
[MATH]9 * \dfrac{8m}{n} - v = \dfrac{p}{n} \implies[/MATH]
[MATH]9u - v = \dfrac{p}{n}.[/MATH]
[MATH]\text {Let } w = 9u - v.[/MATH]
[MATH]\therefore w = \dfrac{p}{n}.[/MATH]
Now how do you finish?

 

[lex]

n|(72m-p) and n|(8m)

Now n|8m [MATH]\rightarrow[/MATH] n|9*8m
i.e n|72m

Therefore n|[72m – (72m-p)]
i.e. n|p

 

[JeffM]

@lex

Elegant, but maybe too elegant for this student at this stage.

But bravo.

 

[lex]

I suppose it has the virtue of simplicity, but as you say perhaps too simple. Nimis simplificandum!

 

[JeffM]

Nimis simplificandum!

Quick. Gerund or supine?

 

[lex]

Gerund! (Memories come flooding back).

 

[JeffM]

Yep

Now to stretch your mind a bit

Lapidus
Das Hase
In gramine
Im Grasse
Sedebat
Er sas
Et edebat
Und as.

These profundities stick with you.

 

[lex]

@JeffM
I think the Roman Empire has been invaded! I suspect this is not Cicero.

 

[JeffM]

@JeffM
I think the Roman Empire has been invaded! I suspect this is not Cicero.

Unknown author. Definitely post Cicero. Tacitus perhaps, but the use of rhyme strongly suggests 8th or 9th century C.E. at the earliest. Not my period so I cannot pin it down more specifically.