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

1. Originally Posted by Evo Guy 911
Dudes I can prove that equation wrong. See this:
n^2 - n -41 = never divisible by 7
To prove it wrong, we have to prove that for a certain value of n, the answer must b divisible by 7. So we work counter-clock wise, we put the equation equal to anything divisible by 7. Now watch:
n^2 - n - 41 = 7
=> n^2 - n = 7+41
=> n(n-1) = 48
Now either n=48, or (n-1)=48. So we put n=48 in the above equation

48^2=2304

48^2 - 48 - 41=7
2215=7 (I know it doesn't sound right, but for the time being.....)

When we divide 2215 by 7, it gives 316.428. Not a whole nmber. Now try the other equation. i.e., n-1=48 => n=47

Putting in equation:

47^2=2209

47^2 - 47 -41=7
2209 - 47 - 41=7
2121=7

This solves to give the answer 303. A perfect whole number. I know nobody asked for a proof, Just for the fun...... Peace
hullo?? n-1=48 => n=47 ???
i think n-1=48 => n=49
so by ur logic
49^2 - 49 -41 = 7
2401 - 49 - 41 = 7
2311 = 7
When we divide 2311 by 7, it gives 330.1428...

now what???

n math induction WONT work also cos we gotta prove its NOT divisble rite?

wht u do is:
suppose n^2 - n -41 IS divisble ...asumption
so n^2 - n - 41 = 7k [where k is a whole number 0,1,2,3...]
since this is true for all values of n, it must be true for n=0
0^2 - 0 - 41 = 7k
-41 = 7k
so k = -5.857... [NOT a whole number so our assumption is false]

just posting so tht mr singh dont get 'jhapped' tomoro in class hehe

2. Tabber
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thnks zing, lol... i thnk most of us realised that evo guy had typed a load of B.ull S.hit... cud u please xplain wht mathematical induction actually is tho...
btw, try this, 5^2 - 5 - 41 and divide by 7... huh, mad????

3. Tabber
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o, btw zing, i sed in the first post, that n is an integer above 0, i think

4. suppose n^2 - n -41 IS divisble ...asumption
so n^2 - n - 41 = 7k [where k is a whole number 0,1,2,3...]
since this is true for all values of n, it must be true for n=1
1^2 - 1 - 41 = 7k
-41 = 7k
so k = -5.857... [NOT a whole number so our assumption is false]

Originally Posted by mr singh
btw, try this, 5^2 - 5 - 41 and divide by 7... huh, mad????
aaaaaaaaaarggggghhhhhhhh

so i gues we've only proven that our assmptn that n^2 - n - 41 is ALWAYS divisible is not true

n its impossible to prove tht its NEVER divisible by 7 cos that obviously not true

in math induction u follow 3 steps:
1. assume sumthin is true for some 'n'
2. prove its also true for 'n+1' using assumption 1
3. prove its true for n=1..... therefor from 1, 2 & 3 its true for all 'n'

5. otoh 5^2 - 5 - 41 divided by 7 is -3.....not a whole number
mebbe v shd assume k is integer?

is ths frm a text book?? which one??

6. Tabber
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mebbe teacher gav us rong equation, or mebbe i jus copied it down rong, either way, i feel so confused... i will find out 2moro, wen my nxt lesson... hopefuly teacher tells, or i getr angry... maybe the rite formula is plus 41, but isnt that one of i think eulers formulas 4 describing prime numbers up 2 n=41?

7. ur rite
euler frmla is n^2 - n + 41
n^2 + n + 41 is legendres

http://mathworld.wolfram.com/Prime-G...olynomial.html

8. Tabber
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although, no formula can produce all primes, thy r all limited

9. Tabber
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it seems that such quadratics can describe a sequences of primes, but not an infinite sequence... this is sort of demonstrated by ulams doodles, whereby the diagonals are made of a numbers which can b represented by a formula, where the diagonal ends, so does the formula

10. nother way 2 look at it is
n^2 - n - 41 [write -n as -3n + 2n and -41 as -6-35]
n^2 - 3n + 2n - 6 -35
= (n-3)(n+2) -35 [since 35 is neway div by 7 so u can ignore it]
=> (n-3)(n+2) shd be div by 7

this is true wenever n-3 = 7,14,21,... or n+2 = 7,14,21,....

thas y it works when n=5 like u said ...

11. Tabber
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mmmm... impressive... this definately deserves reps, i am most impressed, how did u wrk it out?

13. Tabber
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lol... mebbe i jus banged my hed 2 hard wen tryin 2 wrk it out,
ur thing means that the initial conjecture about the equation being not divisible by 7 is wrong... how weird, that means either me or the teacher made a mistake, neway, i will find out 2moro 4 real... thnks 4 all ur hlp... btw, i wsnt able 2 giv u rep, hav 2 spread sum round

14. Originally Posted by Knight_ki_gf
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....
I gotta go take a piss.....

15. Tabber
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principia mathematica needed 200pages to explain that 1+1=2 ...

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