Square Root of a PrimeDate: 07/14/99 at 20:23:13 From: Stephen F. Moran Subject: Irrational Numbers I am trying to grasp the concept of irrational numbers. Suppose p is a prime number. Show that sqrt(p) is irrational. (Please don't use 2 as an example; the textbook I have uses it and I am missing somthing.) Is there a way to demonstrate this graphically? Steve Moran Date: 07/15/99 at 10:13:07 From: Doctor Floor Subject: Re: Irrational Numbers Dear Steve, Thanks for your question. Here is one proof from the Dr. Math archives: http://mathforum.org/dr.math/problems/smothers7.15.98.html Another proof: Let p be a prime number. To prove that sqrt(p) [sqrt means square root] is irrational number, we have to show that it cannot be written as a/b, where a and b are integers. Such a proof has to be done indirectly: 1. We assume that sqrt(p) could be written as a/b for integer a and b; 2. We show that this leads to a contradiction. So something impossible has to be derived from our assumption. The fact that we derive someting impossible from the fact that sqrt(p) = a/b, shows that sqrt(p) = a/b must be false, and we have proven the theorem. We are going to try this: So let's assume that sqrt(p) could be written as a/b for integer a and b. We can derive: sqrt(p) = a/b [sqare both sides] p = a^2/b^2 [multiply by b^2; ^2 means squared] p*b^2 = a^2 a^2 = p*b^2 Now we have two integers in the equation, p*b^2 and a^2. We can factor these integers into primes, and find: a^2 = p^m * {product of primes unequal to p} p*b^2 = p^n * {product of primes unequal to p} Because a^2 is a square, m has to be even. Because b^2 is a square, and "p*" brings an additional factor p, n has to be odd. So m and n are not equal. But if a^2 = p*b^2, then the factorization into primes of these two has to be equal. So, m and n should be equal. We have found a contradiction. And our assumption, that sqrt(p) = a/b for integers a and b, cannot hold. This proves that sqrt(p) is irrational. I hope this helps you to understand. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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