HSC 2015 MX2 Marathon ADVANCED (archive) (1 Viewer)

[simpleetal]

Re: HSC 2015 4U Marathon - Advanced Level

so ur argument is, if g (x)=x (non zero factor) +h (y) =0, then these zeroes occur only when x=0 and h (y) =0? This is clearly not true, and both of yours + soccerball's solutions wouldn't get more than 1 mark in the actual exam!

Lol, to be fair these people haven't encountered situations where mathematical rigor is of great importance... I do have a rigorous solution, which I can post if requested perhaps tomorrow.

For my proof im essentially proving when both sides are zero what values work. And since theres an extra x^3 it wont work any other way unless theyre both zero...
EDIT: Would that be "legit" enough as a proof

You mean whether your solution is 100% valid? No. But I suppose it would take someone like Glitter or Sy or etc. to truly convince you

 

[Drsoccerball]

Re: HSC 2015 4U Marathon - Advanced Level

Lol, to be fair these people haven't encountered situations where mathematical rigor is of great importance... I do have a rigorous solution, which I can post if requested perhaps tomorrow.



You mean whether your solution is 100% valid? No. But I suppose it would take someone like Glitter or Sy or etc. to truly convince you

No a year 1 student could convince me just as long as they told me what i was doing wrong...

 

[Drsoccerball]

Re: HSC 2015 4U Marathon - Advanced Level

And by making a claim that my method is wrong you have to back up your claims with evidence. Ill be happy to accept that my methods wrong but i dont see any error in what i did. Can you point out my mistake please ?

 

[InteGrand]

Re: HSC 2015 4U Marathon - Advanced Level

And by making a claim that my method is wrong you have to back up your claims with evidence. Ill be happy to accept that my methods wrong but i dont see any error in what i did. Can you point out my mistake please ?

I think you assumed that the LHS and RHS in your last equation both had to be 0? This doesn't have to be true in general for an equation to hold.

 

[Drsoccerball]

Re: HSC 2015 4U Marathon - Advanced Level

I think you assumed that the LHS and RHS in your last equation both had to be 0? This doesn't have to be true in general for an equation to hold.

But i stated that it cant be anything else since theres an extra term on the left? Is there a counter example to it?

 

[Chlee1998]

Re: HSC 2015 4U Marathon - Advanced Level

But i stated that it cant be anything else since theres an extra term on the left? Is there a counter example to it?

so ur argument is, for any diaphantine equation with an unequal number of terms, equality is achieved iff all the terms are zero? Or do u mean something else?

 

[Chlee1998]

Re: HSC 2015 4U Marathon - Advanced Level

But i stated that it cant be anything else since theres an extra term on the left? Is there a counter example to it?

so ur saying, that if ur given x +x^2 plus some term =y +y^2 then the only integer solutions are when x=0? just making sure before I post a counterexample

 

[glittergal96]

Re: HSC 2015 4U Marathon - Advanced Level

But i stated that it cant be anything else since theres an extra term on the left? Is there a counter example to it?

Why would the number of "terms" change anything?

f(x)=g(x) is the same equation as f(x)/2+f(x)/2=g(x), although the latter LHS has two "terms".

I write with quotation marks because "terms" is a completely ambiguous word, as the above example shows.

If you are not convinced, state precisely what you are using in your claimed proof that you believe to be true, and I will provide a counterexample to it.

 

[glittergal96]

Re: HSC 2015 4U Marathon - Advanced Level

Anyway, here is my proof.

There are most likely cleaner ways to do it, and maybe even obvious direct improvements to my method, I feel a little slow today.

 

[Drsoccerball]

Re: HSC 2015 4U Marathon - Advanced Level

Why would the number of "terms" change anything?

f(x)=g(x) is the same equation as f(x)/2+f(x)/2=g(x), although the latter LHS has two "terms".

I write with quotation marks because "terms" is a completely ambiguous word, as the above example shows.

If you are not convinced, state precisely what you are using in your claimed proof that you believe to be true, and I will provide a counterexample to it.

But prove to me that g(x)+g(x)=g(x) has a solution where g(x) is not 0

 

[glittergal96]

Re: HSC 2015 4U Marathon - Advanced Level

But prove to me that g(x)+g(x)=g(x) has a solution where g(x) is not 0

Of course it doesn't, but you did not use this precise statement in your proof.

 

[Drsoccerball]

Re: HSC 2015 4U Marathon - Advanced Level

Of course it doesn't, but you did not use this precise statement in your proof.

thats what ive been saying for the last few posts ahahah...

 

[glittergal96]

Re: HSC 2015 4U Marathon - Advanced Level

thats what ive been saying for the last few posts ahahah...

What have you been saying in your last few posts? I have not seen you say the thing you said in the last post earlier?

No-one is disputing that t=2t => t=0, but that doesn't make your proof attempt any more valid...

 

[Chlee1998]

Re: HSC 2015 4U Marathon - Advanced Level

What have you been saying in your last few posts? I have not seen you say the thing you said in the last post earlier?

No-one is disputing that t=2t => t=0, but that doesn't make your proof attempt any more valid...

Lol don't worry about Drsoccerball, he doesn't know what he's talking about/ can't admit that he couldn't solve the question.

But your solution is excellent!

 

[Chlee1998]

Re: HSC 2015 4U Marathon - Advanced Level

thats what ive been saying for the last few posts ahahah...

But prove to me that g(x)+g(x)=g(x) has a solution where g(x) is not 0

But i stated that it cant be anything else since theres an extra term on the left? Is there a counter example to it?

And by making a claim that my method is wrong you have to back up your claims with evidence. Ill be happy to accept that my methods wrong but i dont see any error in what i did. Can you point out my mistake please ?

No a year 1 student could convince me just as long as they told me what i was doing wrong...

For my proof im essentially proving when both sides are zero what values work. And since theres an extra x^3 it wont work any other way unless theyre both zero...
EDIT: Would that be "legit" enough as a proof

Whats wrong with my solution...

[]

What have you been saying in your last few posts? I have not seen you say the thing you said in the last post earlier?

No-one is disputing that t=2t => t=0, but that doesn't make your proof attempt any more valid...

I have compiled ball's relevant posts. No where has he included in his proof that "g(x)+g(x)=g(x) so g(x) = 0."

 

[Sy123]

Re: HSC 2015 4U Marathon - Advanced Level

I have compiled ball's relevant posts. No where has he included in his proof that "g(x)+g(x)=g(x) so g(x) = 0."

What are you doing?

This is a marathon, not a court, post a question.

 

[Librah]

Re: HSC 2015 4U Marathon - Advanced Level

Twenty-one girls and twenty-one boys took part in a mathematical competition. It turned out that each contestant solved at most six problems, and for each pair of a girl and a boy, there was at least one problem that was solved by both the girl and the boy. Show that there is a problem that was solved by at least three girls and at least three boys.

 

[InteGrand]

Re: HSC 2015 4U Marathon - Advanced Level

Twenty-one girls and twenty-one boys took part in a mathematical competition. It turned out that each contestant solved at most six problems, and for each pair of a girl and a boy, there was at least one problem that was solved by both the girl and the boy. Show that there is a problem that was solved by at least three girls and at least three boys.

IMO haha

 

[Sy123]

Re: HSC 2015 4U Marathon - Advanced Level

 

[Drsoccerball]

Re: HSC 2015 4U Marathon - Advanced Level

Twenty-one girls and twenty-one boys took part in a mathematical competition. It turned out that each contestant solved at most six problems, and for each pair of a girl and a boy, there was at least one problem that was solved by both the girl and the boy. Show that there is a problem that was solved by at least three girls and at least three boys.

Does it specify how many questions?