Comments on: Roots of integers http://www.johndcook.com/blog/2009/12/20/roots-of-integers/ The blog of John D. Cook Sat, 11 Feb 2012 01:10:06 -0500 http://wordpress.org/?v=2.8.4 hourly 1 By: Weiwei http://www.johndcook.com/blog/2009/12/20/roots-of-integers/comment-page-1/#comment-123707 Weiwei Tue, 20 Dec 2011 16:26:40 +0000 http://www.johndcook.com/blog/?p=3944#comment-123707 @caio Indeed! Thanks for the clarification :) @caio Indeed! Thanks for the clarification :)

]]> By: (Quadrat-)Wurzeln aus ganzen Zahlen « Flos Mathe-Blog http://www.johndcook.com/blog/2009/12/20/roots-of-integers/comment-page-1/#comment-123704 (Quadrat-)Wurzeln aus ganzen Zahlen « Flos Mathe-Blog Tue, 20 Dec 2011 16:06:57 +0000 http://www.johndcook.com/blog/?p=3944#comment-123704 [...] fast jeder noch aus der eigenen Schulzeit. In dem vor zwei Jahren veröffentlichten Artikel “Roots of integers” in John D. Cook’s Blog (“The Endeavour“) entdeckte ich heute einen viel [...] [...] fast jeder noch aus der eigenen Schulzeit. In dem vor zwei Jahren veröffentlichten Artikel “Roots of integers” in John D. Cook’s Blog (“The Endeavour“) entdeckte ich heute einen viel [...]

]]> By: Caio Braz http://www.johndcook.com/blog/2009/12/20/roots-of-integers/comment-page-1/#comment-123507 Caio Braz Tue, 20 Dec 2011 01:27:53 +0000 http://www.johndcook.com/blog/?p=3944#comment-123507 Weiwei, we've got the two cases covered: Part 1 is ok, if n is an integer, if its square root is an integer, its ok! Part 2: if some integer has a rational (i.e. a number in form a/b, with a and b integers) square root, then, some fraction (a/b)^2 will be this integer. The proof stands that it's impossible to a fraction yield this result, so if the square root is not an integer, it can't be a rational, so, it's irrational Weiwei, we’ve got the two cases covered:
Part 1 is ok, if n is an integer, if its square root is an integer, its ok!

Part 2: if some integer has a rational (i.e. a number in form a/b, with a and b integers) square root, then, some fraction (a/b)^2 will be this integer.

The proof stands that it’s impossible to a fraction yield this result, so if the square root is not an integer, it can’t be a rational, so, it’s irrational

]]> By: Weiwei http://www.johndcook.com/blog/2009/12/20/roots-of-integers/comment-page-1/#comment-123490 Weiwei Tue, 20 Dec 2011 00:18:09 +0000 http://www.johndcook.com/blog/?p=3944#comment-123490 The statement is that an integer is either (1) a perfect square or (2) its square root is <b>irrational</b>. What about the 2nd part? The statement is that an integer is either (1) a perfect square or (2) its square root is irrational. What about the 2nd part?

]]> By: sherifffruitfly http://www.johndcook.com/blog/2009/12/20/roots-of-integers/comment-page-1/#comment-121759 sherifffruitfly Tue, 13 Dec 2011 16:37:04 +0000 http://www.johndcook.com/blog/?p=3944#comment-121759 "Denise, I agree that is seems odd that proofs in high school are limited to geometry." The intellectual horsepower level of education majors doesn't permit anything more than that. “Denise, I agree that is seems odd that proofs in high school are limited to geometry.”

The intellectual horsepower level of education majors doesn’t permit anything more than that.

]]> By: John http://www.johndcook.com/blog/2009/12/20/roots-of-integers/comment-page-1/#comment-30245 John Tue, 05 Jan 2010 16:32:34 +0000 http://www.johndcook.com/blog/?p=3944#comment-30245 Denise, I agree that is seems odd that proofs in high school are limited to geometry. (And from what I understand, proofs have largely been removed from geometry.) Number theory seems like a reasonable place to learn proofs. Denise, I agree that is seems odd that proofs in high school are limited to geometry. (And from what I understand, proofs have largely been removed from geometry.) Number theory seems like a reasonable place to learn proofs.

]]> By: Denise Gaskins http://www.johndcook.com/blog/2009/12/20/roots-of-integers/comment-page-1/#comment-30239 Denise Gaskins Tue, 05 Jan 2010 14:06:29 +0000 http://www.johndcook.com/blog/?p=3944#comment-30239 I remember being amazed when I first saw this proof, at how easy it was. I wondered then (and still do) why we never had it in school. We never did any proofs except in geometry and trig. I remember being amazed when I first saw this proof, at how easy it was. I wondered then (and still do) why we never had it in school. We never did any proofs except in geometry and trig.

]]> By: I. J. Kennedy http://www.johndcook.com/blog/2009/12/20/roots-of-integers/comment-page-1/#comment-29340 I. J. Kennedy Mon, 21 Dec 2009 21:43:07 +0000 http://www.johndcook.com/blog/?p=3944#comment-29340 Perhaps you're right, but I'm not sure. In any case we know <i>a</i>^n and <i>b</i>^n are relatively prime by the <a href='http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic' rel="nofollow">Fundamental Theorem of Arithmetic</a>. If <i>a</i> factors to certain primes, then multiplying <i>a</i> times itself isn't going to yield any new primes. So the factorizations of <i>a</i>^n and <i>b</i>^n have no primes in common, because <i>a</i> and <i>b</i> don't. Perhaps you’re right, but I’m not sure. In any case we know a^n and b^n are relatively prime by the Fundamental Theorem of Arithmetic. If a factors to certain primes, then multiplying a times itself isn’t going to yield any new primes. So the factorizations of a^n and b^n have no primes in common, because a and b don’t.

]]> By: John Armstrong http://www.johndcook.com/blog/2009/12/20/roots-of-integers/comment-page-1/#comment-29337 John Armstrong Mon, 21 Dec 2009 21:09:56 +0000 http://www.johndcook.com/blog/?p=3944#comment-29337 How do you know that an and bn are relatively prime? The work amounts to the messy parts in the post. This version is only simpler because it hides the hard bits in a lemma. How do you know that an and bn are relatively prime? The work amounts to the messy parts in the post. This version is only simpler because it hides the hard bits in a lemma.

]]> By: John http://www.johndcook.com/blog/2009/12/20/roots-of-integers/comment-page-1/#comment-29332 John Mon, 21 Dec 2009 17:02:28 +0000 http://www.johndcook.com/blog/?p=3944#comment-29332 I think your proof is easier to follow than mine. I think your proof is easier to follow than mine.

]]> By: I. J. Kennedy http://www.johndcook.com/blog/2009/12/20/roots-of-integers/comment-page-1/#comment-29330 I. J. Kennedy Mon, 21 Dec 2009 16:57:12 +0000 http://www.johndcook.com/blog/?p=3944#comment-29330 Another way to get the same result is to assume a/b is fraction in lowest terms and is not an integer (i.e. b ≠ 1), and consider powers (a/b)<sup>n</sup>. Clearly a<sup>n</sup> and b<sup>n</sup> are relatively prime, and the denominator b<sup>n</sup> ≠ 1, so a<sup>n</sup>/b<sup>n</sup> is not an integer. In other words, no (non-integer) fraction, when raised to a power, can produce an integer. Another way to get the same result is to assume a/b is fraction in lowest terms and is not an integer (i.e. b ≠ 1), and consider powers (a/b)n. Clearly an and bn are relatively prime, and the denominator bn ≠ 1, so an/bn is not an integer. In other words, no (non-integer) fraction, when raised to a power, can produce an integer.

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