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### Irrational Numbers; Rational Square Roots

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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?
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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/
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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?
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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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