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**JaneFairfax****Member**- Registered: 2007-02-23
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#9(i)

For (ii), there are still gaps.

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**smiyc86****Member**- Registered: 2009-03-19
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well logically it is understood ... ....

I love Maths and Music ... dunno which more

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**JaneFairfax****Member**- Registered: 2007-02-23
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No, it is not. Would you like to attempt #9(ii) again?

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**smiyc86****Member**- Registered: 2009-03-19
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*Last edited by smiyc86 (2009-03-31 22:43:13)*

I love Maths and Music ... dunno which more

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**JaneFairfax****Member**- Registered: 2007-02-23
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No, still wrong.

Also you still have not proved #1 satisfactorily.

*Last edited by JaneFairfax (2009-04-03 21:42:42)*

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**JaneFairfax****Member**- Registered: 2007-02-23
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**smiyc86****Member**- Registered: 2009-03-19
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hahahaha ... u really know just one way to prove a prob and think that it is the only way ... let me tell u dear ... there are more than 5 ways to solve any problem in math

I love Maths and Music ... dunno which more

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**smiyc86****Member**- Registered: 2009-03-19
- Posts: 78

@jane ... prove me 1+1=2

I love Maths and Music ... dunno which more

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**JaneFairfax****Member**- Registered: 2007-02-23
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Of course there many ways to solve a problem. That does not mean that any old way to solve it is correct. Some of your solutions are correct, some are not.

For example, your solution to #6 is correct, even though it is different from mine. My solution uses the AMGM inequality.

If your solution is correct, then congratulations. If it isnt, and I point out that your solution is wrong, then you should go back and check your working instead of being stubborn about it.

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**JaneFairfax****Member**- Registered: 2007-02-23
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I thought I should give the solution to #7 because that problem has to do with algebraic topoolgy but I cant expect every MathsIsFun browser to be familiar with algebraic topology, fascinating though the subject is.

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**smiyc86****Member**- Registered: 2009-03-19
- Posts: 78

@jane ... if my solution is wrong i will accept that before even u can tell me about it .... but if i am right ... u need to accept it ... there are no old ways in math .... u only accept some new ways if u think it is better than the previous one .... but in no way i can find your solution way better than mine ... both of us used contradiction to solve it ... so how can mine be old ... infact u used the old way of squaring both sides etc. the one we used to do in schools ....

so in my books ur style of solution is in no way nearly as interesting or nearly as appealing as mine ...

oh yes ... as u say my solution is wrong ....

tell me one step which is wrong ....:lol:

if u can't show one such step ... never again say someone wrong

*Last edited by smiyc86 (2009-04-03 17:51:23)*

I love Maths and Music ... dunno which more

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**JaneFairfax****Member**- Registered: 2007-02-23
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smiyc86 wrote:

lets assume m√d + n√e is rational ... then we can represent it in the form p/q ... p,q ∈ Z

now there must be two integers a and b such that

p=a+b

and m√d = a/q ; n√e = b/q

>> √d = a/mq ; √e = b/nq ..... as a,b,mq,nq ∈ Z

√d and √e becomes rational numbers ... which contradicts

That is where your solution to #9(ii) wrong.

Your solution to #1 has a flaw as well. But I will leave you to spot it yourself as an exercise.

*Last edited by JaneFairfax (2009-04-03 21:43:02)*

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**smiyc86****Member**- Registered: 2009-03-19
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hmm agreed nice spotting .... thanx.... [:)]

I love Maths and Music ... dunno which more

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**JaneFairfax****Member**- Registered: 2007-02-23
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**JaneFairfax****Member**- Registered: 2007-02-23
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*Last edited by JaneFairfax (2009-04-15 03:43:52)*

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

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

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

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**JaneFairfax****Member**- Registered: 2007-02-23
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By the way, what do people think of my exercise questions? (I set all of them myself.) Are they too difficult? Or too easy?

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**JaneFairfax****Member**- Registered: 2007-02-23
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NB: A *p*-group is a group in which every element has order a power of the fixed prime *p* (including the identity, which has order *p*[sup]0[/sup]). It is worth noting that such a group need not be finite there do exist infinite groups in which every element has finite order.

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**JaneFairfax****Member**- Registered: 2007-02-23
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**JaneFairfax****Member**- Registered: 2007-02-23
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**kean****Member**- Registered: 2009-05-17
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don't think it is good question.

[img]C:\Users\libo\Desktop\New Text Document (2).html[/img] produces

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**JaneFairfax****Member**- Registered: 2007-02-23
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**JaneFairfax****Member**- Registered: 2007-02-23
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