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Artic_counter
11-7-2010, 02:50 PM
So yeah, every Friday since the start of the semester, I've stayed at school with a teacher to prepare for an upcoming math competition (All of Canada) . Among the work I've been given, I'm stuck on this one and I would be very grateful of you if you could explain me how to solve it.

a) : I'm at a loss
bi) : Piss easy
bii) : I'm stuck
biii) : #@!\$%

Edit : Oh and I'm not allowed to use a calculator. Just to let you know in case that changes anything.

http://img703.imageshack.us/img703/5656/maths.jpg

iironiic
11-7-2010, 03:32 PM
Hmm... for the first question... I don't know if this is correct, and I'm not sure if my justification is strong enough to support it, but this is my answer:

1) 42/99
Reason: First I noticed that if you want a fraction to be close to 3/7, you need to add or subtract a fraction relatively close to 0 to 3/7. Generally fractions with larger denominators have smaller increments, and therefore, closer approximations.

So how did I get my answer. First, 3/7 = 42/98. Then what I did was multiply both sides of this equality by 98/99. From this you get: 42/99 = 3/7 (98/99) < 3/7. Or, 42/99 = 3/7 (1-1/99) = 3/7 - 1/(33*7) < 3/7. This means that the positive difference is 1/(33*7). Then considering 43/99, you have to add 1/99 onto 42/99. By doing this, you see that 3/7 < 43/99 = 3/7 - 1/(33*7) + 1/ (33*3) = 3/7 + 1/33 (1/3 - 1/7) = 3/7 + 4/ (3*7*3*11). So the positive difference is 4/(21*33).

So comparing the two, we see that 1/(33*7) < 4/(21*33) because multiplying 33 on both sides get you 1/7 < 4/21 or 3/21 < 4/21. So my answer is 42/99.

There might be some flaws in my thinking, but this is what my answer would've been if I took this test lol

EDIT: I'll try the baseball question now lol
EDIT 2: I am stumped with both questions, but I have a feeling that the third question had something to do with Fibonacci numbers. Odd lol.

The second one.... wtf lol

Artic_counter
11-7-2010, 03:55 PM
hahaha xD "wtf how am I supposed to solve that" is what comes to mind xD

For your answer on a), I've started looking into it but I won't be able to finish it this evening since I'll need to go study for my math exam tomorrow (A feasible one at least xD)

Thanks :D

iironiic
11-7-2010, 03:59 PM
hahaha xD "wtf how am I supposed to solve that" is what comes to mind xD

For your answer on a), I've started looking into it but I won't be able to finish it this evening since I'll need to go study for my math exam tomorrow (A feasible one at least xD)

Hahah xD I literally said the same exact words! :P

The reason why I wasn't sure about my first answer was that I failed to account for every number from 1 - 99 and I made a leap by assuming that you need a large denominator to get really close to 3/7. So it's possible that it's wrong, but I'll let you determine that :)

Good luck studying! I need to finish Latexing my math homework now haha!

Patashu
11-7-2010, 04:21 PM
4a. We want the number closest to 3/7, and the difference made by changing the numbers is minimized when the numbers start out large. Notably, because of the /7, you can add 3/7 to the top and 1 to the bottom as many times as you like and p will never be closer to an integer than 1/7, meaning that all fractions that are adjusted to be close to p/q for small integers will be quite 'far off' because of the amount of adjustment required.
(3/7)*14 = 42/98.
If we subtract this from another nearby fraction, we'll call this p/q.
Any number around this will have ROUGHLY 98^2 for q, so we should look at p.
p will equal (42*otherq)-(otherp*98). Wanting this to be as small as possible, otherq should be nearly 98 and otherp should be nearly 42, and in addition they should have deviations on opposite sides of each other (i.e. one higher, other lower)
So we might try 42/99, 41/99, 41/98, 42/97 and 43/97.
By exhaustion we find that 42/99-42/98 is the closest - 42/(99*98) - while 42/98-42/97 is very close, 42/(98*97).

4b.

0/1 1/1
0/1 1/2 1/1
0/1 1/3 1/2 2/3 1/1
0/1 1/4 1/3 2/5 1/2 3/5 2/3 3/4 1/1
0/1 1/5 1/4 2/7 1/3 3/8 2/5 3/7 1/2 4/7 3/5 5/8 2/3 5/7 3/4 4/5 1/1
0/1 1/6 1/5 2/9 1/4 3/11 2/7 3/10 1/3 4/11 3/8 5/13 2/5 5/12 3/7 4/9 1/2 5/9 4/7 7/12 3/5 8/13 5/8 7/11 2/3 7/10 5/7 8/11 3/4 7/9 4/5 5/6 1/1

Let's look at the sequence of numerators:
0 1
0 1 1
0 1 1 2 1
0 1 1 2 1 3 2 3 1
0 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 1
0 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 1 5 4 7 3 8 5 7 2 7 5 8 3 7 4 5 1

Let's look at the sequence of denominators:
1 1
1 2 1
1 3 2 3 1
1 4 3 5 2 5 3 4 1
1 5 4 7 3 8 5 7 2 7 5 8 3 7 4 5 1
1 6 5 9 4 11 7 10 3 11 8 13 5 12 7 9 2 9 7 12 5 13 8 11 3 10 7 11 4 9 5 6 1

the second column holds 1/1, 1/2, 1/3, 1/4... 1/n
the third column holds 1/n-1
the fourth column holds 2/(n*2-3)
the fifth column holds 1/n-2
the sixth column holds 3/(n*3-7)
...
the fourth last column holds 3/5, 5/7, 7/9... (analog to fourth column)
the third last column holds n-2/n-1
the second last column holds 0/1, 1/2, 2/3, 3/4... n-1/n

maybe work from there?

Artic_counter
11-7-2010, 04:58 PM
I think you're onto something patashu but I've found a way to solve it xD I'm going to try explaining (warning : My native language is french).

P.S.: iff = if and only if
just in case that's not how you say it

We consider two rational numbers a/b and c/d which are next to each other in a particular stage.

This means : a/b < c/d
but also : bc - ad > 0

The rational number that is going to be inserted between a/b and c/d at the next stage is (a+c)/(b+d).

a/b < (a+c)/(b+d) iff a(b + d) < b(a + c) iff bc − ad > 0

which we know is true, also

(a+c)/(b+d) < c/d iff d(a + c) < c(b + d) iff bc − ad >0

which is also true

That gives us

a/b < (a+c)/(b+d) < c/d

and this tells us that every rational numbers inserted at any given stage is between the rational numbers on it's sides. Therefore, every rational number inserted between rational numbers must be strictly larger than the first and strictly smaller than the second. The three numbers are strictly increasing.

Therefore, no rational number will be inserted more than once.

Edit : **** I need to go study.

MrRubix
11-7-2010, 05:02 PM
Answer to the first one is 41/96, btw.

Artic_counter
11-7-2010, 05:05 PM
Answer to the first one is 41/96, btw.

How did you find it ? Would totally love you if you could explain it to me xD

11-7-2010, 05:47 PM
For part (iii):

To prove every rational number will exist, you have to prove:

1. For every denominator x, each numerator from 0 to x (adding 1 each time) exists.
2. Every denominator will eventually be hit
 Uhh 1 is theoretically wrong given part (ii); I meant to say every numerator that gives a unreducible fraction

2 is easy enough, right? If you just keep adding the left or right most baseball sum denominator, you get a sequence of n+1 from 1->infinity.

So for 1, you can use a general case for which every numerator with a given denominator is hit, right?

I don't really feel like spelling out a case using variables but, let's take denominator 5:

0/5 = 0/1
1/5 occurs at stage 4
2/5 occurs at stage 3
3/5 occurs at stage 3
4/5 occurs at stage 4

So it looks like for cases (1/x) and (x-1/x) will take place at stage (y) and every unique case between 1 and x-1 will occur in stage (y-1). Any reducible fractions in a fraction set with a given denominator can just be reduced to a value that has already occurred, by (ii) [right?]

I dunno if that helps or not