What use is there for rationalizing denominator?

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    Algebra Radical
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Discussion Overview

The discussion revolves around the utility of rationalizing denominators in radical expressions, particularly in the context of algebra. Participants explore the reasons for this technique, its implications, and its relevance in various mathematical scenarios.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions the necessity of rationalizing denominators, suggesting that it merely switches the rationality of the numerator and denominator without clear benefit.
  • Another participant proposes that rationalizing can simplify expressions, providing an example of transforming 6 / √12 into √3.
  • A different viewpoint highlights that dividing an irrational number by a rational number is often easier, especially without computational tools.
  • Participants discuss the application of rationalization in fractions involving complex numbers, illustrating this with the example of simplifying 1/i using the complex conjugate.
  • One participant expresses a personal preference for a specific format of writing mathematical expressions, indicating that simplicity in notation can vary among individuals.

Areas of Agreement / Disagreement

Participants express differing opinions on the usefulness and implications of rationalizing denominators, indicating that multiple competing views remain without a clear consensus.

Contextual Notes

Some discussions touch on the formatting of mathematical expressions in the forum, which may not directly relate to the main topic of rationalizing denominators but highlight the technical aspects of communication in mathematics.

leighflix
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I've been learning algebra for the past 2 years (in high school), not once have we ever had to rationalize a denominator in a radical expression. I am now relearning Algebra and Trig., what use is there? I mean, all you're doing is switching the numerator (rational) to the Denominator (irrational) in terms of rationality. (not actually switching the numbers).

Rational / Irrational → Irrational / Rational
 
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After quite a while of thinking about this, the only conclusion I can come up with is to simplify it.

Say for example, 6 / √12 → 6√3 / √(12 ⋅ 3) → 6√3 / 6 → √3
 
leighflix said:
I've been learning algebra for the past 2 years (in high school), not once have we ever had to rationalize a denominator in a radical expression. I am now relearning Algebra and Trig., what use is there? I mean, all you're doing is switching the numerator (rational) to the Denominator (irrational) in terms of rationality. (not actually switching the numbers).

Rational / Irrational → Irrational / Rational
The main rationale behind this technique is that it's easier to divide an irrational number by a rational number than the other way around, especially if you don't have a calculator or computer to do the work for you.

The technique is also used for fractions that involve complex numbers. For example, ##\frac 1 i## can be simplified by multiplying by the complex conjugate over itself; i.e., by multiplying by ##\frac{-i}{-i}## (which is 1). So ##\frac 1 i = \frac 1 i \cdot \frac{-i}{-i} = \frac{-i}{1} = -i##.
 
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Thanks, and a small side question: How do you create the fractions? (on this forum)
 
Regarding the format of math equations: If you see an example math equation here whose format you want to mimic, you can right-click and see the Show Math As => TeX commands. Copy it and surround it with [ tex] ... [/tex]. (Note. I had to insert a space before 'tex' to stop the formatter from parsing it). Here is an example from a post above.

\frac 1 i = \frac 1 i \cdot \frac{-i}{-i} = \frac{-i}{1} = -i
 
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FactChecker said:
Copy it and surround it with [ tex] ... [/tex].
I never use these any more -- I find it simpler to use $$ at the beginning of the expression and $$ at the end -- it's less to type. For stuff that I want to show inline, I use ## at the beginning and two more of them at the end. The ## pairs are equivalent to ....
 
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I personally consider ##\frac{1}{\sqrt{2}} + \frac{1}{\sqrt[3]{3}} = \frac{\sqrt2}{2} + \frac{\sqrt[3]9}{3} = \frac{3\sqrt2+2\sqrt[3]9}{6}## to be simpler than ##\frac{1}{\sqrt2} + \frac1{\sqrt[3]3} = \frac{\sqrt2+\sqrt[3]3}{\sqrt2 \sqrt[3]3}## but this is a YMMV thing.
 

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