Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

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

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 Bézouts 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 Bézouts 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.

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 19,182

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.

Offline

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

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

Im 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 Ive posted my first three exercise questions.

Offline

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

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.

Offline

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

Correct.

Offline

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

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.

Offline

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

Excellent job.

Now someone try #1? Its not that difficult.

Offline

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

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 *n* ≤ *x*, 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)*

Offline

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

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

Offline

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

Youre more or less on the correct line of thought. However the problem is to show that *c* is divisible by *ab*. Its true that *ab* is less than or equal to *c*, but this doesnt show that *c* is divisible by *ab*.

Ive already given a short solution using Bézouts identity.

Offline

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

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

Offline

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

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.

Offline

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

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)*

Offline

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

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.

Offline

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

JaneFairfax wrote:

Youre more or less on the correct line of thought. However the problem is to show that

cis divisible byab. Its true thatabis less than or equal toc, but this doesnt show thatcis divisible byab.Ive already given a short solution using Bézouts identity.

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

Offline

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

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?

Ive read your proof again and I can see what youre 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.

Offline

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

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" .

Offline

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

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)*

Offline

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

Well, Ive 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)*

Offline

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

right?

Numbers are the essence of the Universe

Offline

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

Numbers are the essence of the Universe

Offline

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

Stanley_Marsh wrote:

right?

No. You have the right idea, though.

Stanley_Marsh wrote:

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

Offline

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

Numbers are the essence of the Universe

Offline

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

Thats the way to do it.

Offline

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

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.

Offline