SOLUTION: Hello out there,I need some help with some really tough problems,I only have two of them,the first one is,Determine whether the number is rational or irrational.If it is irrational

Algebra ->  Radicals -> SOLUTION: Hello out there,I need some help with some really tough problems,I only have two of them,the first one is,Determine whether the number is rational or irrational.If it is irrational      Log On


   



Question 113395: Hello out there,I need some help with some really tough problems,I only have two of them,the first one is,Determine whether the number is rational or irrational.If it is irrational,find two consecutive integers between which its graph lies on the number line,-sqrt81.And the last one is,Determine whether sqrt 67 is rational or irrational.If it is irrational,find two consecutive integers between which its graph lies on the number line.Thanks to whoever can help.
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
-sqrt%2881%29=-9 because +9%2A9=81. 9 is clearly a rational number because it can be expressed as the ratio of two integers: 9%2F1 for example.

67 is not a perfect square, because 8%5E2=64 and 9%5E2=81. So, sqrt%2867%29 is some number between 8 and 9. But, is it rational or irrational?

Let's suppose the square root of 67 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 sqrt%2867%29=a%2Fb, so

67=a%5E2%2Fb%5E2
a%5E2=67b%5E2

This means that a squared is a multiple of 67. Now you can easily see
that to get a number that is a multiple of 67 by squaring a, a itself
must be a multiple of 67, so a+=+67c for some whole number c. Putting
this back into the equation, we get

67%5E2c%5E2+=+67b%5E2
Now we divide both sides of the equation by 67, getting
67c%5E2=+b%5E2
So b%5E2 is a multiple of 67, and b must be a multiple of 67.
But now we are in trouble, because we have shown that both a and b are
multiples of 67, but we know that a and b have no common factor, since we chose a%2Fb at the start to be reduced to lowest terms. So
our original assumption that sqrt%2867%29 was rational must
have been wrong. Therefore, sqrt%2867%29 is irrational.

Hope this helps,
John