Rational or Irrational? Featuring a NEW technique!

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Is this crazy product rational or irrational? This comes from the 2022 HMMT from Harvard University.
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Опубликовано:

14 окт 2024

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Комментарии : 15

@joseluishablutzelaceijas928 2 месяца назад

This is indeed quite an interesting solution, thanks for sharing this.

@JPiMaths 2 месяца назад

@@joseluishablutzelaceijas928 no worries, glad you enjoyed it!

@TheMichaelmorad 2 месяца назад

this trick made me smile

@JPiMaths 2 месяца назад

@@TheMichaelmorad I'm glad!! It's a neat little trick, right?!

@theflaggeddragon9472 2 месяца назад

Wonderful proof! I tried to imitate the proof that e is irrational to no avail; writing the number as a/b and after you get you your sum, write it as sum(+/-) n^(k_i) where k_i is an increasing sequence. Then multiplying by bn^(k_n), the first n terms give integers, and the rest of the sum has absolute value less than 1, so you get that a difference of integers has absolute value less than 1. But the sum is alternating and plausibly give zero which cannot be ruled out.

@JPiMaths 2 месяца назад

@@theflaggeddragon9472 ah nice attempt. It's often not a bad idea to try and mimic other proofs

@winniedobrokot 2 месяца назад

You can use a system for writing numbers with a non-fixed base, unlike base 10, where each digit position has a unique base: use 2 as the base for the first digit, 3 for the second, 4 for the third digit, and so on. The usual base-10 system for the sequence 0,b1 b2 b3 b4… gives the number as the sum of b_i/10^i. Our number system with variable bases will represent the number as the sum of b_i/(2*3*4*…*i). Conversion from a number to our new system is similar to the base-10 conversion. To get each digit, we divide the interval into 10 equal sub-intervals and determine which sub-interval contains our number. We then set this sub-interval as the new interval and repeat the process. In the new number system, the number of sub-intervals is not fixed; it is 2 for the first digit, 3 for the second digit, and so on. The sequence 0,1234… is forbidden in our system in the same way that 0,9999… is forbidden in base 10. Our new number system for fractional parts has the nice property that any rational number has a finite representation. However, e is not finite; it is represented as 2,11111… in this system.

@cauchym9883 2 месяца назад

A beautiful argument. Now I wonder, of course, is the number you get even transcendental? Looks like Liouville's theorem should be applicable here.

@JPiMaths 2 месяца назад

@@cauchym9883 potentially... I've not studied too much into transcendental numbers. I just looked up Liouville's Theorem but I could only find the complex numbers one; what's the theorem to do with transcendental numbers?

@cauchym9883 2 месяца назад

@@JPiMaths Ah, sorry. Liouville has many theorems named after him. If you look up "Liouville numbers" on Wikipedia, you'll find the reference I've meant.

@copernicus633 2 месяца назад

What is the fallacy?: Since each term in the product is rational, and products of rationals are rational, the infinite product is rational.

@hassanalihusseini1717 2 месяца назад

It would be tue for finite many factors, but not necessarily for an infinite number.

@JPiMaths 2 месяца назад

@@copernicus633 yep, just because something is true for finitely n, doesn't mean it's true in the limit. For example if x_n=π/n then x_n is irrational for each n but lim x_n=0 which is rational