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/