Wikipedia:Reference desk/Archives/Mathematics/2007 October 7

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October 7

French Franc 1965

Shall appreciate if anyone can tell me today's value in Euros of 15,000 1965 French Francs. Info needed to illustrate history of a small village.86.197.151.8 08:52, 7 October 2007 (UTC)petitmichel

I have moved this question to the Humanities section, here.  --Lambiam 12:57, 7 October 2007 (UTC)

Thanks. Just for interest the equivalent sum in 2006 (old francs) would have been approx. 115,500 francs! Appreciation of 100,000! Whow! (Yes, this is a crude figure, but sufficient for my needs.86.200.4.62 14:33, 7 October 2007 (UTC)petitmichel

Proof question

Hello. I just did a question that I would like someone to check for me. It would be much appreciated.

The positive integers can be split into five distinct arithmetic progressions, as shown:

A: 1, 6, 11, 16
B: 2, 7, 12, 17
C: 3, 8, 13, 18
D: 4, 9, 14, 19
E: 5, 10, 15, 20

a) Write down an expression for the value of the general term in each progression.
b) Hence prove that the sum of any term in B and any term in C is a term in E.
c) Prove that the square of every term in B is a term in D.
d) State and prove a similar claim about the square of every term in C.

a)

$A=5n-4$

$B=5n-3$

$C=5n-2$

$D=5n-1$

$E=5n$

b)

${\begin{aligned}B+C&=(5m-3)+(5m-2)\\&=(5m+5m)+(-2-3)\\&=10m-5\\&=5(2m-1)\\\end{aligned}}$

Since this number is multiple of 5, it is a term in E.

c)

${\begin{aligned}B^{2}&=(5n-3)^{2}\\&=25n^{2}-30n+9\\\end{aligned}}$

${\begin{aligned}25n^{2}&\equiv 0{\text{ (mod }}5)\\-30n&\equiv 0{\text{ (mod }}5)\\9&\equiv 4{\text{ (mod }}5)\\\implies 25n^{2}-30n+9&\equiv 4{\text{ (mod }}5)\\\end{aligned}}$

${\begin{aligned}D&=5n-1\\5n&\equiv 0{\text{ (mod }}5)\\-1&\equiv 4{\text{ (mod }}5)\\\implies 5n-1&\equiv 4{\text{ (mod }}5)\\\end{aligned}}$

Therefore since both are congruent to 4 (mod 5) it is true.

d)

${\begin{aligned}C^{2}&=(5n-2)^{2}\\&=25n^{2}-20n+4\\\end{aligned}}$

${\begin{aligned}25n^{2}&\equiv 0{\text{ (mod }}5)\\-20n&\equiv 0{\text{ (mod }}5)\\4&\equiv 4{\text{ (mod }}5)\\\implies 25n^{2}-20n+4&\equiv 4{\text{ (mod }}5)\\\end{aligned}}$

So, as proved above, the square of any term in C is also a term in D. —Preceding unsigned comment added by 172.200.78.35 (talk) 15:08, 7 October 2007 (UTC)

a) and b) are good. b) would be even better if you remark that $2m-1$ must be a positive integer (for example, $5(-1)=-5$ is not in E, and you need to mention that this doesn't happen here).

In c) and d), you have the right idea, but the way you have written it doesn't really prove the needed result. You say "every square of an element of B is congruent to 4 modulo 5, and every element of D is congruent to 4 modulo 5, hence every square of an element of B is is an element of D", which is a fallacy (for example, every man breathes, and every woman breathes, but not every man is a woman). What you should have shown is the converse, that every number congruent to 4 modulo 5 is an element of D. However, I think this just complicates matters; it is easier to just say that $(5n-3)^{2}=5(5n^{2}-6n+2)-1$ and show that $5n^{2}-6n+2$ is a positive integer, and hence this is in D. The same goes for part d). -- Meni Rosenfeld (talk) 15:25, 7 October 2007 (UTC)

Thank you Meni, I'll take that on board. 172.200.78.35 15:33, 7 October 2007 (UTC)

Just an after thought, would my problem with b) and your solution to c) and d) have been solved if I said n>o? 172.200.78.35 15:38, 7 October 2007 (UTC)

First, let me correct myself; there is a relatively serious error in your solution to b). You have an element of B and an element of C; You know that the former is some multiple of 5 minus 3, and and latter is some multiple of 5 minus 2. But it doesn't have to be the same multiple. So your numbers are not $5m-3$ and $5m-2$ but rather $5n-3$ and $5m-2$, where n and m are positive integers which may or may not be the same.

Now, stating that n (and m) is positive is just the start. From a) we learn that:

• If n is a positive integer, then $5n-4$ is in A, and so on;
• If some number is in A, then there is some positive integer $n$ such that the number is equal to $5n-4$, and so on.

So a correct solution to b) would be: Let x be in B and y be in C. Then there are some positive integers n and m such that $x=5n-3$ and $y=5m-2$. So $x+y=5(m+n-1)$, and $m+n-1$ is a positive integer. Therefore, $x+y$ is in E. -- Meni Rosenfeld (talk) 16:15, 7 October 2007 (UTC)