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## #1 2007-03-05 16:39:49

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Janes exercises

Two integers are said to be relatively prime iff their highest common factor is 1. Thus, 2 and 5 are relatively prime, as are 4 and 9.

A result called Bézouts identity states that if a and b are nonzero integers that are relatively prime, there exist integers x and y such that ax + by = 1. {Note that x and y can be negative as well as positive.)

1. Consider the number 60. 60 is divisible by 4 and 5, and 60 is also divisible by 4×5 = 20. However, 60 is divisible by 4 and 10 but not by 4×10 = 40. 4 and 5 are relatively prime, whereas 4 and 10 are not.

Using Bézouts identity (or otherwise) prove that if an integer c is divisible by both a and b, where a and b are nonzero, relatively prime integers, then c is also divisible by the product ab.

2. Prove that the product two consecutive even numbers is divisible by 8.

3. Hence (or otherwise) prove that if n is an odd integer that is not divisible by 3, n[sup]2[/sup]−1 is divisible by 24.
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## #2 2007-03-05 16:45:34

ganesh
Registered: 2005-06-28
Posts: 25,023

### Re: Janes exercises

The one easiest to answer is #2, since any two consecutive even numbers are such that one of them is certainly divisible by 2. Hence, the product would be divisible by 8.

Nice questions, JaneFairfax, I suggest you name the topics based on the subjects posted and suffix or prefix them Jane-1 etc. depending on the number of the post!

Your posting in the Exercises section is much appreciated!!!

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #3 2007-03-05 17:09:02

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

Well, youre on the right track with your answer to #2, but you need to elaborate just a little bit.

Im intending to use this thread to post some interesting exercise questions on various math topics (and various math levels); some of the questions will be interrelated. Thus Ive posted my first three exercise questions.

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## #4 2007-03-05 17:37:10

pi man
Member
Registered: 2006-07-06
Posts: 251

### Re: Janes exercises

An even number can be expressed as 2x.   The next even number would be 2x + 2.   The product would be 4x^2 + 4x = 4x( x+ 1).   Either the x or the (x+1) has to be even and therefore divisible by 2.  Combine that with the  4, and you have your 8.

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## #5 2007-03-06 03:58:29

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Correct.

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## #6 2007-03-06 04:23:52

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: Janes exercises

n² -1 can be factorised into (n+1)(n-1). We are given that n is odd, and so n+1 and n-1 are both even. Hence, by the answer to 2), n² -1 is divisible by 8.

We are also told that n is not divisible by 3. This means that it takes either the form 3k+1 or 3k+2, where k is an integer.

For the case of 3k+1, this would mean that n-1 was equal to 3k and so divisible by 3.
For the case of 3k+2, this would mean that n+1 was equal to 3k+3 and so divisble by 3.

Therefore, one of (n+1) and (n-1) is always divisible by 3 and so n²-1 is always divisible by 3.

We have previously shown that n²-1 was also divisible by 8, and combining these pieces of information shows that n²-1 is divisble by 24.

Why did the vector cross the road?
It wanted to be normal.

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## #7 2007-03-06 08:39:16

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

Excellent job.

Now someone try #1? Its not that difficult.

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## #8 2007-03-10 18:13:00

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

More exercises from me.

4. If x is real, what is 0[sup]x[/sup]?

5. If x is a real number, [x] denotes the greatest integer less than or equal x. Prove that for any real numbers x and y,

In other words, the greatest-integer function is a superadditive function. (Hint: If n is any integer such that nx, then n ≤ [x].)

6. A cycloid is the curve traced by a fixed point on a circle rolling along a straight line. If the radius of the generating circle is r, the parametric equations of the cycloid are

Find the area bounded by one arch of the generated curve (0 ≤ θ ≤ 2π) and the x-axis.

Last edited by JaneFairfax (2007-03-10 18:18:22)

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## #9 2007-03-19 05:11:53

Kurre
Member
Registered: 2006-07-18
Posts: 280

### Re: Janes exercises

1. every number that isnt a prime can be factorized in its prime factors, for example 36 can be factorized in 2*2*3*3.
so the number c can be factorized in the prime factors P1*P2*P3.....Pn
a and b are relatively prime so they cant have one same prime factor, but since c is divisible by both a and b, they must be created by one or some of the prime factors in c. therefor a*b can never be more than c.

for example a can be P1*P2 and b can be P3*P4. if we divide c with a and b there will still be P5*P6*....*Pn left.
but if a is P1*P2, b cant be for example P2*P3 since then both will have P2 as a prime factor.

would this be a correct proof?? still practising

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## #10 2007-03-19 12:12:54

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

Youre more or less on the correct line of thought. However the problem is to show that c is divisible by ab. Its true that ab is less than or equal to c, but this doesnt show that c is divisible by ab.

Ive already given a short solution using Bézouts identity.

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## #11 2007-03-19 13:45:04

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

Last edited by JaneFairfax (2007-03-19 19:19:07)

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## #12 2007-03-19 13:56:22

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: Janes exercises

I thought that 0[sup]0[/sup] was indeterminate? There's a problem because 0^n = 0, but n^0 = 1, so there we have a contradiction.

Why did the vector cross the road?
It wanted to be normal.

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## #13 2007-03-19 19:18:33

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

0[sup]0[/sup] is defined as 1. Try it on a calculator.

And

so the graph is not right-continuous at x = 0.

Last edited by JaneFairfax (2007-03-19 19:22:26)

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## #14 2007-03-19 22:09:12

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: Janes exercises

What calculator are you using? Any that I try just tell me Ma Error or something similar.

Why did the vector cross the road?
It wanted to be normal.

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## #15 2007-03-19 23:18:26

Kurre
Member
Registered: 2006-07-18
Posts: 280

### Re: Janes exercises

JaneFairfax wrote:

Youre more or less on the correct line of thought. However the problem is to show that c is divisible by ab. Its true that ab is less than or equal to c, but this doesnt show that c is divisible by ab.

Ive already given a short solution using Bézouts identity.

well since ab is created by the prime factors in c, c must be divisible by ab?

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## #16 2007-03-20 13:49:50

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

mathsyperson wrote:

What calculator are you using?

A decent one.

Kurre wrote:

well since ab is created by the prime factors in c, c must be divisible by ab?

Ive read your proof again and I can see what youre trying to get at  so, yes.

Proofs involving writing out long sequences of primes can sometimes become blurred with details and hard to follow  I generally avoid them if I can find alternative proofs.

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## #17 2007-03-20 15:26:23

Zhylliolom
Real Member
Registered: 2005-09-05
Posts: 412

### Re: Janes exercises

Letting 0[sup]0[/sup] be 1 is a standard convention, which helps eliminate special cases where some theorems break down. There are some combinatorial and set-theoretic justifications for this convention. However, in analysis 0[sup]0[/sup] is treated as an indeterminate. Moreover, consider x[sup]x[/sup] = e[sup]x ln x[/sup]. There is no real number corresponding to ln 0, so x[sup]x[/sup] is not defined at x = 0; in other words 0[sup]0[/sup] is undefined. The calculator says it is equal to 1 because of the common convention mentioned above; but a real decent calculator would notify you something like "Warning: 0^0 replaced by 1" .

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## #18 2007-04-03 03:12:37

Kurre
Member
Registered: 2006-07-18
Posts: 280

### Re: Janes exercises

Thanks for the tip Jane.
I really liked exercise 2 and 3, i really need to practise proving these types of problems. i would really appreciate if you could create more exercises like them

Last edited by Kurre (2007-04-03 03:12:56)

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## #19 2007-04-03 04:22:38

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

Well, Ive been trying to create some more exercises for this thread. Here is something I just this moment made up.

7. Let f(x) be a periodic odd function with period n. In other words, f(x) satisfies f(x) = −f(x) (odd function) and f(x+n) = f(x) (periodic with period n). If n is an odd integer greater than 1, prove that

8. Use the above result to show that

NB: For #8 you must use the result in #7.

Last edited by JaneFairfax (2007-04-16 10:27:46)

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## #20 2007-04-04 11:12:50

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

### Re: Janes exercises

right?

Numbers are the essence of the Universe

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## #21 2007-04-04 11:15:36

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

### Re: Janes exercises

Numbers are the essence of the Universe

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## #22 2007-04-04 20:13:19

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

Stanley_Marsh wrote:

right?

No. You have the right idea, though.

Stanley_Marsh wrote:

True, but as Ive stated, I want you to use the result in #7 to do #8.

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## #23 2007-04-05 02:29:10

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

### Re: Janes exercises

Numbers are the essence of the Universe

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## #24 2007-04-09 00:17:12

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

Thats the way to do it.

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## #25 2007-04-11 17:08:01

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: Janes exercises

9. Prove that a monotone-increasing sequence of real numbers that is bounded above converges to its least upper bound.

10. Give an example of a monotone-increasing sequence of rational numbers that is bounded above which does not converge to a rational number.

11. Show that nevertheless a monotone-increasing sequence of rational numbers that is bounded above is a Cauchy sequence.

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