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This question already has an answer here:

While solving a problem I came to an answer $\frac{1}{\sqrt{3}+1}$.

But this was not the answer.

The answer was $\frac{\sqrt{3}-1}{2}$ which comes on rationalisation of my answer.

Then I divided both results and get the same decimal value.

Then why do we need rationalisation if it doesn't change decimal value?

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marked as duplicate by Solomonoff's Secret, Shailesh, Adam Hughes, Namaste, C. Falcon Jan 12 '17 at 0:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Sometimes you need to take the natural log of such an answer. Many times, rationalizing an answer as to to having a monomial in the denominator makes it easy to work with the rules of logarithms i.e. expanding and condensing. $\endgroup$ – imranfat Jan 11 '17 at 19:18
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In the pre-calculator days, it was a handy skill. If you need to compute $2/\sqrt{3}$ then you have to do long division of $2$ by $1.732.$ Ick. But if you rationalize the denominator, you get $2(1.732)/3$. It's much easier to divide by $3$ than $1.732.$

But even in these days were you can just ask Siri for the answer, it's still a handy skill. It's easier to compare sizes of numbers if the denominators are rational. Some integrands are easier to integrate if the radicals are moved to the numerator. And probably some other things I can't think of.

So asking students in pre-calc to put their answers in this standard form is preparing them for later stuff.

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We don't need it. In fact your answer is correct. However, it is standard practice to rationalize those type of fractions. This makes it easier to compare result; it's like having some sort of "standard representation" of a number instead of several, much in the same way as when we write $1/3$ and not $27/81$

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    $\begingroup$ Indeed there has been more than one question on this site about why answers are different, when in fact they are the same. Having a standard form is just a convention. This also mirrors the procedure with complex numbers where denominators are converted to real numbers to get the form $a+bi$. Further it helps with adding such numbers when they are components of an answer rather than the whole answer. When algebraic expressions are involved this procedure can help to identify the magnitude of an expression where limits are involved. $\endgroup$ – Mark Bennet Jan 11 '17 at 18:45
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It's generally easier to deal with fractions where the numerator is complicated; for instance, $\frac{\sqrt{3}-1}{2}=\frac{\sqrt{3}}{2}-\frac{1}{2}$, but there's no similar obvious way to split up $\frac{1}{\sqrt{3}+1}$.

As a result, it's often more useful to rationalize. Before calculators, when it was hard to check if two expressions represented the same number, it was common to always rationalize. These days it's less common to insist on it, but being able to do it when needed is important: situations like that come up a lot in calculus, and rationalizing at the right time can make a problem much simpler, or even be the difference between a problem looking unsolvable, and turning it into something easy. (Of course, in calculus the fractions usually have variables, so it's harder for a calculator to check if they're equal.)

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