Question 190858
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You can't rationalize *[tex \LARGE  \frac{8}{sqrt{32}}], because no matter what you do, you will still have an irrational factor somewhere in the expression.  What you <i><b>can</b></i> do is rationalize <u>the denominator</u>, which is what I suspect you meant all along.  The language of mathematics is a very precise thing; please endeavor to use it correctly.  By the way, if your teacher/instructor actually taught you to say it that way, <i><b>shame</b></i> on him or her -- and you have my permission to quote me, including the emphasis. Having vented my spleen, let's get on with it.


First realize that *[tex \LARGE  sqrt{32} = sqrt{16\,\cdot\,2} = 4 sqrt{2}], so:


The expression becomes: *[tex \LARGE  \frac{8}{sqrt{32}} =  \frac{8}{4 sqrt{2}} = \frac{2}{sqrt{2}}]


So far, so good, but we still have that pesky *[tex \LARGE sqrt{2}] in the denominator.  The only thing that will change an irrational square root into a rational number is multiplying the root by itself.  However, you can't just multiply the denominator by something other than 1 without changing the value of the denominator and therefore the entire fraction.  The solution is to multiply your fraction by 1, but in the form of *[tex \LARGE \frac {sqrt{2}} {sqrt{2}}], thus:


*[tex \LARGE \left(\frac{2}{sqrt{2}}\right)\left( \frac {sqrt{2}} {sqrt{2}}\right) = \frac{2 sqrt{2}}{2} = sqrt{2}]


If you are asking yourself why I didn't just divide 2 by root 2 to get root 2 without all of the folderol, it is because these sorts of problems very seldom come out this nicely.  It is essential that you learn the process, without any shortcuts, from the very beginning.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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