Math From Scratch 15: The Fundamental Theorem of Arithmetic

The fifteenth lesson of the Math From Scratch series, dealing with the Fundamental Theorem of Arithmetic, is available here.

4 replies on “Math From Scratch 15: The Fundamental Theorem of Arithmetic”

I’m enjoying this series. I’m a bit confused by the following statement though

As an example, let us prove that if p is prime, and if a1, a2, …, an are integers, then there exists at least one i for which p divides ai.

Am I missing some other constraint on these variables? What if I pick p = 7, n = 3, a1 = 1, a2 = 2, and a3 = 3?

No, I missed listing the constraint. It should read that, if p divides the PRODUCT of all n integers, then p must divide one of the integers on the list. I’ll get that corrected and reposted this week.

While on the topic of corrections, should “the result follows from the assumption of the n + 1 case” refer to the case for “n” and not “n + 1”?

Thanks for catching those. Both errors should now be corrected.

I’m enjoying this series. I’m a bit confused by the following statement though

Am I missing some other constraint on these variables? What if I pick p = 7, n = 3, a1 = 1, a2 = 2, and a3 = 3?

No, I missed listing the constraint. It should read that, if p divides the PRODUCT of all n integers, then p must divide one of the integers on the list. I’ll get that corrected and reposted this week.

While on the topic of corrections, should “the result follows from the assumption of the n + 1 case” refer to the case for “n” and not “n + 1”?

Thanks for catching those. Both errors should now be corrected.