Hello Brilliant community! I have here some questions about definitions of mathematical concepts. While most of these do not involve problem-solving or proving (as indicated by the title), you might find some of these challenging - or trivial, depending on your perspective.

- Why do we have to rationalize the denominator of a fraction if it contains a radical?
- Does the definition of a kite require it to be a convex polygon?
- Are all parallelograms trapezoids?
- Why is \( \frac {0} {0} \) indeterminate, but \( \frac {1} {0} \) undefined? What is the difference between the two terms? (Wanna go the extra mile? Answer this question without using limits or other calculus concepts.)
- Trapezoid and trapezium - one has a pair of parallel sides, the other has none. Which is which?

More questions to come. Post your comments below, and see if people agree with your answers. Happy solving! Or proving. Answering. Whatever. Cheers.

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TopNewest5) "Trapezium" is the messed up one, there isn't an universal agreement on what it means. It can mean a quadrilateral with two parallel sides or with no pair of parallel sides. Everybody agrees that "Trapezoid" means a quadrilateral with two parallel sides. – Michael Mendrin · 2 years, 1 month ago

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– Francis Gerard Magtibay · 2 years, 1 month ago

I thought this was a matter of differences between American and British English, because Wolfram said so. Thanks anyway, Mr. Mendrin. :)Log in to reply

– Michael Mendrin · 2 years, 1 month ago

We all know that the British drive on the wrong side of the road too.Log in to reply

– Francis Gerard Magtibay · 2 years, 1 month ago

This just made my day. :) No offense, British-English speakers. :)Log in to reply

4) \(\frac { 0 }{ 0 } \) may have multiple finite values, while \(\frac { 1 }{ 0 } \) cannot be expressed by any finite value, as \(0\) times any finite value is \(0\), a contradiction. – Michael Mendrin · 2 years, 1 month ago

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2) I leave this one to the reader as an exercise.

Edit: Okay, since nobody's answering this one, a kite is a quadrilateral that has two pairs of sides of equal length, but with each such pair of equal length sides sharing a common vertex. Contrast this with a parallelogram, where no such pair of equal length sides share a common vertex. This doesn't preclude a kite that is non-convex, i.e., it can vaguely look like a Chevron or maybe an paleolithic arrowhead. – Michael Mendrin · 2 years, 1 month ago

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3) Yes – Michael Mendrin · 2 years, 1 month ago

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trapezoid. Some say it hasexactlyone pair of parallel sides, while according to others, it hasat leastone such pair of sides. So does it just depend on which definition we use? Or is there something wrong with one of them? – Francis Gerard Magtibay · 2 years, 1 month agoLog in to reply

1) Who says we have to rationalize the denominator of a fraction if it contains a radical (or anything else for that matter)? The only justification for it would be that it could lead to wrong results of one doesn't, but it won't. It just makes further manipulation or numerical computation easier if we do--but not necessarily always easier. It depends. – Michael Mendrin · 2 years, 1 month ago

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It's simply a matter of preference!

It's much easier to add two fractions with different denominators if these happen to be integers rather than radicals. For instance, \( \frac {\sqrt{2}}{2} + \frac {\sqrt{3}}{3} \) is much easier to evaluate than its equivalent \( \frac {1}{\sqrt{2}} + \frac {1}{\sqrt{3}} \).

Rational numbers are defined as numbers of the form \( \frac {p}{q} \) where \( p, q \in \mathbb {Z} \) and \( q \neq 0 \). It follows that the denominator of a rational number is always a nonzero integer, If we multiply a rational number \( \frac {p}{q} \) by a radical \( r \), the product \( \frac {pr}{q} \) will always contain a radical in the

numerator, not thedenominator.The first two are a bit related, and the third might have some logical loopholes. Feel free to comment! – Francis Gerard Magtibay · 2 years, 1 month ago

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– Michael Mendrin · 2 years, 1 month ago

There's a distinction between "have to" and "ought to". Yes, we ought to rationalize the denominator, but at the end, it doesn't lead to wrong results if we don't bother to.Log in to reply

– Francis Gerard Magtibay · 2 years, 1 month ago

Indeed.Log in to reply