## Rationalizing Denominators

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### Rationalizing Denominators

by NYC493 » Fri Mar 23, 2012 8:47 am
Hi everyone,

I'm not so good at simplifying complicated expressions. I see the the OG 12th Edition makes a few references to the method of "rationalizing the denominator". An example of this is OG Problem Solving #117, answer explanation on pg. 223:

#117) If n is positive, which of the following is equal to 1/sqrt(n+1) - sqrt(n)?

OG goes on to explain that the denominator can be rationalized by multiplying the expression by sqrt(n+1) + sqrt(n)/ sqrt(n+1) + sqrt(n).

Can anyone explain in plain english what it means to "rationalize the denominator", and why it's useful? In the case of #117, why is it achieved by multiplying sqrt(n+1) - sqrt(n) by sqrt(n+1) + sqrt(n)? Seems like there's a trick that I don't know about.

Thanks!!!

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by GMATGuruNY » Fri Mar 23, 2012 9:23 am
NYC493 wrote:Hi everyone,

I'm not so good at simplifying complicated expressions. I see the the OG 12th Edition makes a few references to the method of "rationalizing the denominator". An example of this is OG Problem Solving #117, answer explanation on pg. 223:

#117) If n is positive, which of the following is equal to 1/sqrt(n+1) - sqrt(n)?

OG goes on to explain that the denominator can be rationalized by multiplying the expression by sqrt(n+1) + sqrt(n)/ sqrt(n+1) + sqrt(n).

Can anyone explain in plain english what it means to "rationalize the denominator", and why it's useful? In the case of #117, why is it achieved by multiplying sqrt(n+1) - sqrt(n) by sqrt(n+1) + sqrt(n)? Seems like there's a trick that I don't know about.

Thanks!!!
To RATIONALIZE a fraction is to remove a âˆš from the denominator.
To remove a âˆš from the denominator, we take advantage of the DIFFERENCE OF TWO SQUARES:
(x+y)(x-y) = xÂ² - yÂ².

(x+y) and (x-y) are CONJUGATES of each other.
If either (x+y) or (x-y) is in the denominator and has a âˆš, the âˆš can be removed by multiplying by the conjugate.
In order not to change the value of the fraction, we multiply BOTH THE TOP AND THE BOTTOM by the conjugate.
(Multiplying the top and the bottom by the same value is the same as multiplying by 1, leaving the value of the fraction unchanged.)

In the problem at hand, we must rationalize 1 / (âˆš(n+1) - âˆšn).
The conjugate of (âˆš(n+1) - âˆšn) is (âˆš(n+1) + âˆšn).
Multiplying both the top and the bottom by the conjugate, we get:

1 / (âˆš(n+1) - âˆšn)

= 1 * (âˆš(n+1) + âˆšn) / (âˆš(n+1) - âˆšn)(âˆš(n+1) + âˆšn)

= (âˆš(n+1) + âˆšn) / (âˆšn+1)Â² - (âˆšn)Â²

= (âˆš(n+1) + âˆšn) / (n+1) - n

= (âˆš(n+1) + âˆšn) / 1

= âˆš(n+1) + âˆšn.

I offered an alternate solution to this problem here:

https://www.beatthegmat.com/tough-ps-t94365.html
Mitch Hunt
Private Tutor for the GMAT and GRE
[email protected]

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