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daisy

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Help in Math Induction Question

prove that for every non-negative integer n , n^3 mod 6 = n mod 6.

(use mathematical induction)

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 46,182

Re: Help in Math Induction Question

Put n=7.
7mod6=1
7^3mod6=343mod6=1
Put n=8
8mod6=2
8^3mod6=512mod6=2.
Assume this is true for k.
Therefore, kmod6=k^3mod6.
Try for k+1.
Lets say (k+1)mod6=m
(k+1)^3mod6 = (k^3 + 3k^2+3k+1)mod6.
= (k^3+3k^2+2k+k+1)mod6= (k^3+2k^2+k^2+2k+k+1)mod6
= [k(k^2+2)+k(k+2)+k+1]mod6...
Running out of time...gotta leave....

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It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Help in Math Induction Question

P(n+1) is (n^3 + 3n^2 + 3n + 1)%6 = (n+1)%6 
and assume we start with n^3%6 = n%6, where % means mod (like in C language).
(n^3 + 3n^2 + 2n)%6 + (n+1)%6 = (n+1)%6
(n^3 + 3n^2 + 2n)%6 = 0
Substitute n%6 in place of n^3%6 and get:
(3n(n+1))%6=0
For even numbers, 3n is a multiple of 6, so that works.
For odd numbers, the (n+1) part is even, so that works.

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Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 46,182

Re: Help in Math Induction Question

Put n=7.
7mod6=1
7^3mod6=343mod6=1
Put n=8
8mod6=2
8^3mod6=512mod6=2.
Assume this is true for k.
Therefore, kmod6=k^3mod6.
Try for k+1.
Lets say (k+1)mod6=m
(k+1)^3mod6 = (k^3 + 3k^2+3k+1)mod6.
= (k^3+3k^2+2k+k+1)mod6= (k^3+2k^2+k^2+2k+k+1)mod6

= {[k²(k+2)+k(k+2)]+k+1}mod6
= {[(k²+k)(k+2)] +k+1}mod6
={[(k(k+1)(k+2) + k+1}mod6

We know that k(k+1)(k+2) is divisble by 6 for any k>1, k∈N,
Hence the above is reduced to
(k+1)mod6

It is seen that it is true for k+1, hence, it is true for any value of k.

q.e.d

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It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.