Prove by induction if n is a natural number and a and b are integers and a=b(modn) then a^m=b^m(mod n)? - m&b mod respitory
show for every natural number m, induction.
Equal sign = means congruent.
Friday, February 12, 2010
M&b Mod Respitory Prove By Induction If N Is A Natural Number And A And B Are Integers And A=b(modn) Then A^m=b^m(mod N)?
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2 comments:
Evidence of induction generally bore me, are also listed. Ok here goes: Base case (m = 1) is clear from a mod n = b induction hypothesis: k = ^ k ^ b mod n. We want a ^ (k +1 show) = b ^ (k +1) mod n , thus establishing the induction step and complete the proof by induction. a ^ (k +1) = (a ^ k) (a) = (b ^ k) (a) mod n by induction. (b ^ k) (a) = (b ^ k) (b) = b ^ (k +1) mod n to the baseline. Thus we see that a ^ (k +1 shown), b = n ^ (k +1) mod have
For a test, not based on induction, try this:
Suppose that a = b (mod n). Then ab = k * n is an integer k.
at - b ^ m = (ab) (a ^ (m-1) + a ^ (m-2) b ^ 1 + ... + a 1 b ^ (m-2) + b ^ (m -1 ))
= K * N * (integer)
This shows that m ^ - b = 0 (mod n) or feet to the point:
m = a ^ b ^ (n mod m).
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Nor is this use of induction, as we wanted, but I have someone who will essentially an examination of the induction. This only shows that we do not use induction on this test.
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