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Well, I will be posting a lot of questions in this thread one by one, after completely clearing my doubts in each of them.

Question1:

There are two urns each containing an arbitrary number of balls. Both are non-empty to begin

with. We are allowed two types of operations:

(a) remove an equal number of balls simultaneously from the urns, and

(b) double the number of balls in any one of them.

Show that after performing these operations finitely many times, both the urns can be made

empty.

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**scientia****Member**- Registered: 2009-11-13
- Posts: 224

*Last edited by scientia (2012-12-02 15:31:30)*

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Thank You ---- That was a nice one

Question 2:

If

Prove that (a+b) is a perfect square

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,981

Hi Agnishom;

This problem appears in texts on Diophantine equations and also on the internet but I do not remember where.

All the solutions to

are of the form:

that is proved by expanding using algebra.

Substituting integers into k, m and n will give every solution.

So

If you make k equal to one then

Since they have no factors in common the GCD ( m (m+n), n (m+n), mn) = 1 and so GCD(a,b,c) is 1.

So a + b is a square for every solution. We are done.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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I understood but how about the proof of that?

By the way, I saw something like that here

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,981

Hi Agnishom;

That is my proof. Not firm enough? Lacking in rigor? Or just out and out silly?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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No I like your proof. But how do I prove this:

???

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,981

Hi;

Like with a trig identity I am going to make one side look like the other. Just algebra will be required.

Multiply through by k:

Combine the LHS.

After division by ( m + n ) in the numerator and denominator.

So

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,378

eeekkk! bobbym

I think you've been hacked!

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**scientia****Member**- Registered: 2009-11-13
- Posts: 224

Bobbym is only trying to link to an image but the server on which the image is hosted isn't allowing remote linking without your first visiting the site.

Try this:

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,981

Hi;

I am seeing the image in my post just like yours. Apparently you all are seeing something else. I will upload the image manually in my post.

Scientia;

Thanks for the link. With your permission I am using that.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**scientia****Member**- Registered: 2009-11-13
- Posts: 224

Agnishom is seeing this image:

That was what I initially saw as well. That's because neither of us had visited the site http://a1star.com/ first. Once I went to the site, I didn't see the remote-linking error image any more.

Some sites which restrict remote linking work like this. The problem is that you can see the correct image because you've visited the site, but others who haven't visited the site see a different image. They have to visit the site in order to see the correct image.

bobbym wrote:

Scientia;

Thanks for the link. With your permission I am using that.

You're welcome.

*Last edited by scientia (2012-12-03 01:23:03)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,981

Hi Scientia;

I did not know any of that. Thanks for telling me. Luckily, they did not put something extremely offensive up there.

I apologize to everyone who got confused.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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You got me wrong...

I mean how can you say that all solutions of 1/a + 1/b = 1/c can be put into that form?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,378

You are easy to forgive. Had me worried there for a moment.

Thought you had gone completely screwy.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,981

Hi Bob;

Thought you had gone completely screwy.

Not yet. I honestly was at a bonafide site that I googled for that gif. It seems no matter how smart I think I am and how long I have lived there is still a billion tricks I have never even heard of.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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How can you say that every solution can be expressed in that form?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,981

Hi Agnishom;

If this is true and it appears so

then we can say a = k m (m+n) and b = k n (m+n) and c = k m n where k,m, and n are arbitrary integers that do not equal 0. Also, ( m + n ) does not equal 0. Of course this does not apply here because everything is positive.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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But how do you get that form???

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,981

Hi;

That is a good question. Also, it is a different question. In the book I am looking at he derives it. I just copied it down in my notes long ago because it was a useful relation for solving equations like that.

Generally, when you are doing a proof you use lots of other theorems and axioms. Just knowing they are true is enough, you do not have to derive them.

For instance, if you use the Pythagorean theorem in your proof you do not have to show how it was derived.

I have shown that the above relation is true for the positive integers. That is enough for the original question.

But since you are asking for the derivation, I can copy it down for you. I do not think I understand it well enough to explain it. But you might be right, if the author put it in to the proof there is a good chance it is mandatory. Let me know what you think about that.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**Avon****Member**- Registered: 2007-06-28
- Posts: 80

I agree with bobbym that you don't need to reprove a result every time you want to use it, but if someone wants to use a result that you have never seen before and doesn't provide a proof of it then it is reasonable to be unhappy with the result, as Agnishom is.

Suppose that

where a, b, c are positive integers such that gcd(a,b,c) = 1.Let l = gcd(a,b), m = gcd(a,c), n = gcd(b,c)

Since gcd(a,b,c) = 1, it follows that gcd(l,m) = gcd(l,n) = gcd(m,n) = 1 and so there exist positive integers a', b', c' such that

a = lma' b = lnb' c = mnc'

Also gcd(a', b') = gcd(a', c') = gcd(b', c') = gcd(a', n) = gcd(b', m) = gcd(c', l) = 1.

Now if we substitute our expressions for a, b, c into the original equation we obtain

Since gcd(c', la'b') = 1, it follows that c' = 1, and hence ma' +nb' = la'b'.

Also gcd(a', nb') = gcd(b', ma') = 1, so a' = b' = 1 and so l = m + n.

Therefore a = m(m+n), b = n(m+n), c = mn.

Of course we also have that gcd(m,n) = 1, but we don't need to use this to see that a + b = (m + n)^2.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,981

Hi;

Thanks for posting the derivation but I did not ever think Agnishom's questions were unreasonable. What I did say is that the derivation is not necessary for the proof.

bobbym wrote:

But since you are asking for the derivation, I can copy it down for you. I do not think I understand it well enough to explain it. But you might be right, if the author put it in to the proof there is a good chance it is mandatory. Let me know what you think about that.

I did say I would provide the derivation of this

if he still wanted to see it.

What I am saying is it is an identity for positive integers. This is proved through the algebra in post #8. Where it comes from is not really important. If I had guessed at it and then proved it that would be fine. I could use it as a theorem in Agnishom's question.

Perhaps post#8 is not rigorous enough, but if it is then it holds for all k,m and n that are positive integers because it is an identity. ( Actually it holds for a great deal more but we only need the positive integers. )

To generate all the solutions just means having to iterate k, m and n for 1,2,3,4,5.... sort of like a loop inside a loop inside a loop.

3 nested loops, a structure Agnishom is familiar with I am sure.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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Thank You, Avon

bobbym wrote:

I did say I would provide the derivation of this, if he still wanted to see it.

Yes I wanted to see it, and Avon's post has helped.

These proofs are quite hard to go through and memorise

Let me post Question 3:

It is probably more simple than the previous ones.**A censusman on duty visited a house which the lady inmates declined to reveal their individualages, but said we do not mind giving you the sum of the ages of any two ladies you maychoose. Thereupon the censusman said In that case please give me the sum of the ages ofevery possible pair of you. The gave the sums as follows : 30, 33, 41, 58, 66, 69. The censusmantook these figures and happily went away.**

What are the individual ages of each of them?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,981

Hi;

Call there ages a,b,c,d.

There are four girls needed to make 6 pairs.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**Avon****Member**- Registered: 2007-06-28
- Posts: 80

bobbym, I apologise if I have caused any offence but I think you are missing something.

Here are two statements:

(1) If k,m,n are positive integers and a = km(m+n), b = kn(m+n), c = kmn, then 1/a + 1/b = 1/c.

(2) If a,b,c are positive integers and 1/a + 1/b = 1/c, then there exist positive integers k,m,n such that a = km(m+n), b = kn(m+n), c = kmn.

In post #8 you demonstrate that (1) is true, but in order to solve Agnishom's problem you are using that (2) is true. Hopefully it is clear that (1) does not imply (2).

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