# Thread: 4 the maths lovers...??? help!!!

1. Originally Posted by Knight_ki_gf
well bjr... in proofs thats how you u go about it...

u assume either of the two to be equal to 48,... if you are able to prove the equation for any one ...then the solution holds good for that particular value...

I know it well what i m supporting !!
what??? ????????????????????
n(n-1)= 48 so n is 48???? ....what maths is this?

2. Beginner
Join Date
Mar 2005
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say x = 7
i.e. (x-7) = 0

but if gng about the proof we dont know if x is +ve..
so if x was -ve then x-7 would come to -14
in this case x+7 will be 0

so always the equation (x-7 )(x+7) will lead to 0

now i m pretty sure my approach is okayyy...

^ Do you know the proof of 1+1=2 (just out of curiousity)??
The proof starts from the Peano Postulates (now dont ask me what it is!!!), which define the natural numbers N.

N is the smallest set satisfying these postulates:
P1. 1 is in N.
P2. If x is in N, then its "successor" x' is in N.
P3. There is no x such that x' = 1. ( as dat would make x = 0... hence not a natural number )
P4. If x isn't 1, then there is a y in N such that y' = x.
P5. If S is a subset of N, 1 is in S, and the implication
(x in S => x' in S) holds, then S = N.

Let a and b be in N.
If b = 1, then define a + b = a' (using P1 and P2).
If b isn't 1, then let c' = b, with c in N (using P4),
and define a + b = (a + c)'.

Then we define 2:
2 = 1' ( that is succesor of 1... same as x')

2 is in N by P1, P2, and the definition of 2.

Theorem:

1 + 1 = 2

Proof:

Use the first part of the definition of + with a = b = 1.
Then 1 + 1 = 1' = 2 Q.E.D.

happy????

Yaar.. i m on study leave...dont make me use my brain cells so much....

4. Originally Posted by dennis
what??? ????????????????????
n(n-1)= 48 so n is 48???? ....what maths is this?
Dodo!!

n=48 didnt hold good ... but (n-1)= 48 holds good...
hence the equation n^2-n-41 is divisible by 7 if n = 47

5. Originally Posted by can_i_play
say x = 7
i.e. (x-7) = 0

but if gng about the proof we dont know if x is +ve..
so if x was -ve then x-7 would come to -14
in this case x+7 will be 0

so always the equation (x-7 )(x+7) will lead to 0

now i m pretty sure my approach is okayyy...

i m not too sure

let me get back on this one....

6. Beginner
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Originally Posted by Knight_ki_gf
Dodo!!

n=48 didnt hold good ... but (n-1)= 48 holds good...
hence the equation n^2-n-41 is divisible by 7 if n = 47

thats what he is trying u tell ..
u cannot assume
n-1 = 48

or n = 48

this is where u r gng wrong...

coz there can be various other factors to 48
48 = 1* 48
= 6*8
= 24*2
etc te c...

7. naah...she is right, its actually a little diffferent.
its not 7 in the rhs but 7k.
so u get
n(n-1) = 41+7k
and u can put n=47 for a value of k=303

wow...knight_ki_gf, u have wowed me

8. Thank you....

me feeling drowsy now... done so much maths after..... like 3-4 years !!

9. Beginner
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Originally Posted by dennis
naah...she is right, its actually a little diffferent.
its not 7 in the rhs but 7k.
so u get
n(n-1) = 41+7k
and u can put n=47 for a value of k=303

wow...knight_ki_gf, u have wowed me
sorry dude cant agree with you...
where does this "7k" comes into picture

n(n-1) = 48

we cannot say n or n-1 = 48

it is one of the possible solutions..
ne

but
when we have abc = 0
either one has to be 0

so now can u get the difference ..

**din me taree ***h rahe hain**
no mre math for today...

10. Tabber
Join Date
Apr 2005
Location
england
Posts
242
ok, sorry guys i took so long 2 respond...
just want 2 say thnks 2 every1 4 tryin 2 help me out, lol...
i like the way wen initially everyone sez mathematical induction is the rite way no 1 really explained it... (dont mean 2 sound rude... i no beggars cant b choosers)
to sahrukh i think it was, i am doin further pure 1, in june some time
to can_i_play: i cannot really follow ur proof, perhaps i am going about it the wrong way, or mayb i a bit on the slow side 2day,
@ alpha: the three theorums u asked 4 hav incredibly long solutions... and as is the case in most mathmatical proofs, few if any people other than the author bother 2 check the proof. if computers are used 2 check, it is often doubted. the only mathmatician who wud have checked is paul erdos, but he is dead... may his memory liv on

11. can_i_play...to be a factor of 7 the lhs needent be only 7, it can be 14,28,etc etc
so u put it as 7k and not 7

12. @singh .. FP1!! wow not many people go for Furthers .. .. edexcel? ..i did that last year .. P5 (FP2) is frigin hard .. just watch out for that .. .
.. and the chapter of proof isnt very well written .. i agree .. get the solution book from the library ..i did .. it helped a lot .. best of luck !

13. Tabber
Join Date
Apr 2005
Location
england
Posts
242
i doin the OCR board i think... jus had S1 n C1 the other day, started work on FP1 yesterday, its hard

14. @gf: I didnt understand somethings (only SOME though) but anyway. I bow my brain cells to you.

15. Tabber
Join Date
Apr 2005
Location
england
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i think dennis has managed 2 get it in a simple uderstandable way... in post number 24 i think... thanks dennis... but i bet there is another way which is really long, lol

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