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A Binary Operation Puzzle – Thoughts Your Selections

Due to Boomii from Taiwan for the suggestion!

Suppose the binary operation ☆ is outlined in order that

xy = (x + y) / (1 + xy).

Simplify the expression

{ … [(2 ☆ 3) ☆ 4] ☆ … } ☆ 2021

As normal, watch the video for an answer.

Binary operation puzzle

Or preserve studying.

“All will probably be nicely should you use your thoughts in your choices, and thoughts solely your choices.” Since 2007, I’ve devoted my life to sharing the enjoyment of sport idea and arithmetic. MindYourDecisions now has over 1,000 free articles with no adverts because of group assist! Assist out and get early entry to posts with a pledge on Patreon.


Reply To A Binary Operation Puzzle

(Just about all posts are transcribed shortly after I make the movies for them–please let me know if there are any typos/errors and I’ll appropriate them, thanks).

Let’s work out a number of examples.

2 ☆ 3 = (2 + 3)/(1 + 2·3) = 5/7
(2 ☆ 3) ☆ 4 = (5/7 + 4)/(1 + (5/7)·4) = 9/11
[(2 ☆ 3) ☆ 4] ☆ 5 = (9/11 + 5)/(1 + (9/11)·5) = 7/8 = 14/16
{[(2 ☆ 3) ☆ 4] ☆ 5} ☆ 6 = (7/8 + 6)/(1 + (7/8)·6) = 11/10 = 22/20

Let an be the expression with the most important integer n. Then from the calculations, we will hypothesize a method:

if n is odd, then
an = (n(n + 1)/2 – 1)/(n(n + 1)/2 + 1)
if n is even, then
an = (n(n + 1)/2 + 1)/(n(n + 1)/2 – 1)

Assuming the method is true, we’ve:

= (2021(2021 + 1)/2 – 1)/(2021(2021 + 1)/2 + 1)
= 1,021,615/1,021,616

Proof of method

Let’s use mathematical induction. The method is true for base instances n = 3, 4. Now assume the method is true for an integer okay ≥3. We have to present the method holds for okay + 1. We are going to break the proof into instances the place okay is odd and even.

If okay is odd, then okay + 1 is even and we’ve:

aokay + 1 = aokay ☆ (okay + 1)

By the induction speculation we will substitute for aokay.

= [(k(k + 1)/2 – 1)/(k(k + 1)/2 + 1)] ☆ (okay + 1)
= [(k(k + 1) – 2)/(k(k + 1) + 2)] ☆ (okay + 1)
= [(k(k + 1) – 2)/(k(k + 1) + 2) + (k + 1)]/[1 + (k(k + 1) – 2)/(k(k + 1) + 2)(k + 1)]
= [(k(k + 1) – 2) + (k + 1)(k(k + 1) + 2)]/[(k(k + 1) + 2) + (k(k + 1) – 2)(k + 1)]
= (okay3 + 3okay2 + 4okay)/(okay3 + 3okay2)
= (okay2 + 3okay + 4)/(okay2 + 3okay)
= ((okay + 1)(okay + 2) + 2)/((okay + 1)(okay + 2) – 2)
= ((okay + 1)(okay + 2)/2 + 1)/((okay + 1)(okay + 2)/2 – 1)

This verifies the method holds for okay + 1.

The proof for okay is even is analogous (particulars omitted).

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