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The prime factorization including both existence and uniqueness. I have totally no idea about this problem except the basecase.

In this problem we only consider number greater or equal to 2. So the basecase should be n=2, and it is easy to prove. But for I.H, I have no idea, should I make I.H be n = ${p_1p_2...p_i}$ or sort of things? Is there any hints for me?

Thank you.

SonicFancy
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1 Answers1

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I'll show the existence part, which is where the induction is used. For this proof, it is easiest to use strong induction: you don't simply assume the case $n-1$ in order to prove $n$, but you assume all cases less than $n$. Using this, the proof is rather simple:

The case $n=2$ is our base case, which is obvious. Now let $n$ be any natural number greater than $2$, and assume for our induction hypothesis that a prime factorization exists for every $1<m<n$. If $n$ is prime, then we're done. Otherwise, $n=ab$ where $a,b>1$. Use our induction hypothesis to complete the proof.

Hayden
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  • So uniqueness can't be proved by induction right? – SonicFancy Mar 26 '14 at 00:25
  • I've not seen a proof of uniqueness using induction. I've always seen it done using Euclid's Lemma (which is the underlying reason that there is uniqueness). I'll look around to see if there is a proof that makes use of induction – Hayden Mar 26 '14 at 00:27
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    I found here (http://www2.imperial.ac.uk/~buzzard/maths/research/notes/inductive_proof_of_fundamental_theorem_of_arithmetic.pdf) a proof of uniqueness using induction. It's not too hard to follow, so hopefully this helps on that front. – Hayden Mar 26 '14 at 00:36
  • @SonicFancy See this answer and this one for a complete inductive proof based on ideas of Zermelo. – Bill Dubuque Mar 26 '14 at 00:47