Irrational Numbers; Rational Square Roots
Date: 02/04/98 at 12:03:27 From: Pat McQuatty and Carolyn Gould Subject: Irrational numbers A student wants to know how he can tell that root 10 is irrational. His calculator can only display 10 digits. How can he know whether this is a terminating or repeating decimal, or an irrational number?
Date: 02/04/98 at 15:31:23 From: Doctor Wilkinson Subject: Re: Irrational numbers You could have the most powerful computer in the world, and you still couldn't tell whether a number was rational or irrational by looking at its digits. We know that the square root of 10 is irrational because we can prove it. Here's how: We use what is called an indirect proof. This means we assume the opposite, that the square root of 10 is rational. Based on that assumption, we draw a chain of conclusions until we arrive at a conclusion that just can't be true. When this happens, we know the original assumption must have been false; in this case, we know that the square root of 10 is irrational. (We haven't done this yet! I'm just telling you the strategy we're going to use). So suppose the square root of 10 is a rational number. Then it must be of the form a/b, where a and b are whole numbers. Furthermore, we can assume that a and b don't have any common factor (otherwise we could just divide the numerator and denominator by the common factor). So we have a/b is the square root of 10, so a squared / b squared is 10, or a squared = 10 times b squared. This means that a squared is a multiple of 10. Now you can easily see that to get a number that is a multiple of 10 by squaring a, a itself must be a multiple of 10, so a = 10c for some whole number c. Putting this back into the equation, we get 10 squared times c squared = 10 times b squared Now we divide both sides of the equation by 10, getting 10 times c square = b squared So b squared is a multiple of 10, and b must be a multiple of 10. But now we are in trouble, because we have shown that both a and b are multiples of 10, but we know that a and b have no common factor. So our original assumption that the square root of 10 was rational must have been wrong. -Doctor Wilkinson, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 02/06/98 at 11:19:15 From: Pat McQuatty Subject: Irrational numbers Another question: We know that if a positive integer is a perfect square (like 36) its square root is a positive integer. If a positive integer is not a perfect square (like 38) is its square root always an irrational number, or are some square roots rational?
Date: 02/06/98 at 12:50:05 From: Doctor Sam Subject: Re: irrational numbers Pat, Great question, Pat. The answer is that ONLY perfect squares have rational square roots. Here's why. If a number, like 38 for example, had a rational square root we could write SQRT(38) = a/b and you can assume that a/b has been reduced as a fraction to lowest terms so that they do not have any common factors (other than 1 of course). Now if you square this you get 38 = (a/b)^2 that is (a/b)*(a/b) = 38. But how can the product of two fractions equal a whole number? Only if the fraction can be reduced. And that would mean that a and b would have to have common factors. By the way, your question generalizes nicely: the only whole numbers with rational square roots are the perfect squares; the only whole numbers with rational cube roots are the perfect cubes (1, 8, 27, 64...) and so on. -Doctor Sam, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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