4 the maths lovers...??? help!!!

Discussion in 'The ChitChat Lounge' started by mr singh, Jan 17, 2006.

  1. Varshita

    Varshita New Member

    oh ohk.... got ur point !!
     
  2. shsnawada

    shsnawada Cyborgs & Pasta

    ^ Do you know the proof of 1+1=2 (just out of curiousity)??
     
  3. can_i_play

    can_i_play New Member

    ..

    but now i m confused have i used it correctly...
    :think: :think:
     
  4. can_i_play

    can_i_play New Member


    it is assumed bcoz of some reason....

    and reason is the product is 0 in such cases not else...

    coz xy =0
    can only happpen if either is 0..
     
  5. Varshita

    Varshita New Member

    explanation of (x-7)(x+7) was good... but what i m wondering is why have you used it there??

    wats the context??
     
  6. dennis

    dennis The Bhangra King

    what??? ????????????????????
    n(n-1)= 48 so n is 48???? ....what maths is this?
     
  7. can_i_play

    can_i_play New Member

    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...
     
  8. Varshita

    Varshita New Member

    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.

    Then we define addition recursively:
    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....
     
    vini and shsnawada like this.
  9. Varshita

    Varshita New Member

    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
     
  10. Varshita

    Varshita New Member

    :think: :think:

    i m not too sure

    let me get back on this one....
     
  11. can_i_play

    can_i_play New Member


    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...
     
  12. dennis

    dennis The Bhangra King

    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
     
  13. Varshita

    Varshita New Member

    Thank you....

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

    can_i_play New Member

    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 dikh rahe hain**
    no mre math for today...
     
  15. mr singh

    mr singh New Member

    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
     
  16. dennis

    dennis The Bhangra King

    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
     
  17. shak

    shak Harrr!

    @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 !
     
  18. mr singh

    mr singh New Member

    i doin the OCR board i think... jus had S1 n C1 the other day, started work on FP1 yesterday, its hard
     
  19. shsnawada

    shsnawada Cyborgs & Pasta

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

    mr singh New Member

    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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