Wikipedia:Reference desk/Archives/Mathematics/2007 October 7
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[edit] October 7
[edit] 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
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
[edit] 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)
Since this number is multiple of 5, it is a term in E.
c)
Therefore since both are congruent to 4 (mod 5) it is true.
d)
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(5n2 − 6n + 2) − 1 and show that 5n2 − 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)
