Talk:Composite Number has Prime Factor not Greater Than its Square Root

Should this be moved to Composite Number Has Prime Factor Less Than Or Equal To Its Square Root? --Cynic (talk) 00:27, 6 April 2009 (UTC)

Good call. The title was so long to begin with that I guess I was a bit lazy. Thanks. -Rob 01:02, 6 April 2009 (UTC)

"No Greater Than" would be shorter ... --Matt Westwood 18:28, 6 April 2009 (UTC)

I'm not sure about "$$n = p_1 \times p_2 \times...\times p_j$$" - seems a bit woolly to me. If we need at this stage to invoke the prime decomposition of $$n$$ (and I'm not sure it needs to be at this stage of the definition of the theorem), might it be better to link directly to Prime Decomposition?

I also think we need to add the constraint that $$n \ge 2$$ (for a start, because negative integers don't have real square roots). In fact specifying $$n$$ as being composite means we can also get away with saying $$n \ge 0$$ as the smallest positive composite number is $$4$$. Thoughts? --Matt Westwood 21:35, 12 April 2009 (UTC)

Technically, we defined both prime and composite numbers to be positive integers (correctly, it doesn't make sense to have negative primes by the fundamental theorem of arithmetic, and composite numbers should parallel the definition of primes). So really, we could do whatever people want with what we include. Personally, I think $$n \geq 0$$ makes the most sense, so I'll switch it to that for now. Agreed on referencing prime decomposition though. --Cynic (talk) 03:15, 15 April 2009 (UTC)

Some expositions of number theory do include negative numbers in the definitions of primes. Some go deeper - "Number Theory" by Hasse takes it further than that and extends the definitions to general integral domains. As for composite numbers, they are defined here for all integers, not just positive ones. --Matt Westwood 05:21, 15 April 2009 (UTC)

Another point is that I am uneasy about using $$\N$$ to mean the positive integers. From a purist point of view, $$\Z$$ is isomorphic to an inverse-completion of $$\N$$ but are conceptually completely different animals. I'm not sure at this stage how much it matters, but I have tended to consider that number-theoretic statements ought to be defined on "positive integers" rather than "natural numbers". Don't know whether to be rigorous about it. --Matt Westwood 05:28, 15 April 2009 (UTC)

"An integer greater than 1 which is not prime is defined as composite." From Definition:Prime Number, so you might want to fix that to include negative numbers. I will note that both Wolfram and Wikipedia define primes and composite numbers on the positive integers [I don't have a number theory book I can look at :(] --Cynic (talk) 21:37, 15 April 2009 (UTC)

I might need reminding to get to it - I'm whacked, it's been a long day (and it's 11 pm where I am, I've been up since 6). --Matt Westwood 22:00, 15 April 2009 (UTC)

Here, it is assumed that both a and b are greater than root of 'n'. does it not affect the generality?--Shankara (talk) 09:48, 28 January 2011 (UTC)


 * I don't think so. The proof is that at least one prime factor is less than or equal to square root. What is being done here is to try and derive a contradiction by assuming the existence of two factors both greater than root n that multiplied together make n. What is being demonstrated is that for any two numbers multiplied together to get n, at least one has to be less than or equal to the square root. Where do you think this affects the generality? --prime mover 18:40, 28 January 2011 (UTC) (PS please: 1. add your post to the bottom, and 2. sign it with two hyphens and four tildes, see the box 4th from right on the edit tools above.)