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Math Help - Proof by induction

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    Proof by induction

    I know how to prove that the product n consecutive integers always has a factor of n! but i have found a problem that requires a proof of this by induction, any hints ?
    Last edited by bobak; January 24th 2008 at 03:44 PM.
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    Quote Originally Posted by bobak View Post
    I know how to prove that the product n consecutive integers always has a factor of n! but i have found a problem that requires a proof of this by induction, any hints ?
    Instead of writing, a,a+1,a+2,... we will write a_1,a_2,a_3,... because the notation is nicer. We will prove that if a_1,\cdots ,a_n are consecutive integers then n! divides a_1\cdot \cdots \cdot a_n. If n=2,3 then the proof is straightforward, this will be the base case an we will use induction on it. We assume that if we have any k consecutive integers then k! divides their product and prove that (k+1)! divides any k+1 consecutive integers. In the proof we will use a simple result: a_1\cdot \cdots a_{k+1} \equiv a_2 \cdot \cdots \cdot a_{k+2} (\bmod (k+1)!), let A = a_2\cdot \cdots \cdot a_{k+1} then by induction k! divides A while a_1 \equiv a_{k+2} (\bmod k+1) because a_{k+2} = a_1 + (k+1), thus, a_1 \cdot A \equiv A\cdot a_{k+2}(\bmod (k+1)!) and we proven this result. This result is simply saying that a product of k+1 consecutive integers is divisible by (k+1)! if and only if the product of k+1 integers shifted one integer over is divisible by (k+1)!. We this we can finally prove this result. Remember we need to prove that (k+1)! divides a_1\cdot \cdots \cdot a_{k+1}. If k+1 divides a_1 then k! divides a_2\cdot \cdots a_{k+1} (by induction) and so (k+1)! = (k+1)k! divides a_1\cdot \cdots \cdot a_{k+1}. If k+1 does not divide a_2 then shift the sequence over enough times until the first term of the newly formed sequence is divisible by k+1 and repeat the argument. Q.E.D.
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