Mediant Inequality I

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This is a short, animated visual proof demonstrating what one might called the mediant inequality.
If you like this video, consider subscribing to the channel or consider buying me a coffee: www.buymeacoffee.com/VisualProofs. Thanks!
This animation is based on a visual proof by Richard A. Gibbs from the June 1990 issue of Mathematics Magazine (doi.org/10.2307/2691137 - page 172).
#mathshorts #mathvideo #math #numbertheory #mtbos #manim #animation #theorem #pww #proofwithoutwords #visualproof #proof #iteachmath #fractions #mediant #inequality #mathematics #fareyfractions #farey
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Опубликовано:

1 окт 2024

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Комментарии : 28

@claudiovijaya 8 дней назад

Unbelievable.....power of computer animation...I never understood maths since 1986 math classes but now I see how simple was.. thanks for your content.

@rcb3921 14 дней назад

A very nice visualization of the idea

@Rage_Battle 14 дней назад

Science is all about visualization ur doing the right job bro 🤜🤜

@HoussamZeddou-i7g 13 дней назад

Thanks

@MathVisualProofs 13 дней назад

No problem

@agytjax 5 дней назад

From the triangle shown, (a+c)/(b+d) >= c/d. Maybe the triangle is not drawn correctly. However, the issue with this visualization is, how did we assume that the slope a/b

@MathVisualProofs 5 дней назад

Those ratios are slopes of the hypotenuses. You can see that the hypotenuses have different steepness and those are the inequalities. This does start with assumption that a/b is less than c/d so that they fit as they do in the outer triangle.

@dougr.2398 6 дней назад

Or, define a Mediant function M(a, b, c, d)

@aza-jo9ru 14 дней назад

I think it works only if all the numbers are positive. (Of course a and c may be zero)

@rogiertp 14 дней назад

b = zero will lead to division by 0

@MathVisualProofs 14 дней назад

Yes for sure. Should have said positive rationals.

@MathVisualProofs 14 дней назад

@@rogiertpthey probably meant a and c can be zero :)

@rogiertp 14 дней назад

@@MathVisualProofs fair!

@aza-jo9ru 14 дней назад

thank you i ment a and c

@blah97202 14 дней назад

I don't think this operation is well-defined. 3/6=1/2 but 3/6+1/5 isn't equal to 1/2+1/5.

@MathVisualProofs 14 дней назад

Yes. The mediant depends on two fractions, not really on rational numbers. So you take two integer fractions and produce another. Still, the mediant operation is an interesting one (just not on rational numbers :) ).

@frendlyleaf6187 14 дней назад

yeah, it's a bit finicky cause you gotta take both of the fractions in their lowest terms, that's kinda how it's defined so that it can make some sense and be useful somewhere.

@Ninja20704 13 дней назад

There are already rules in place that we only allow fractions of the simplest form to be used under this operation so that the answer is unique.

@Ninja20704 14 дней назад

Also known as naive fraction addition (since is the “wrong” way to add fractions when you first learn it 😂) That aside, the mediant operation and the inequality are useful for finding rational approximations of irrational numbers using a method known as the Farey algorithm. One small caveat is that you can only use the operation when the two fractions are in their simplest form i.e. irreducible in order to obtain a unique output from the operation