• M0 M1 M3 M5 M6

    Module 0 Day 14 Challenge Part 2

    I think you can prove this pattern algebraically: here are my thoughts. (1/n-1) - 1/n where n and n-1 are consecutive = 1/ (n-1) n. The LCD(Least Common Denominator) is (n-1)n since consecutive numbers are coprime. then, setting the two fractions to their common denominators you get [n-(n-1)] / (n-1)n = 1/(n-1)n therfore, DONE

  • MOD

    @angeliccanary Yep, that's exactly how you would rigorously prove that a pattern exists!

    Just a really small note- for the proof, it actually doesn't strictly matter that n(n-1) is the LCD; we can still algebraically set n(n-1) to be the common denominator no matter what. This is sort of like how we can write 1/2 + 3/5 as 22/20, even though technically it's 11/10 simplified. You're right that n(n-1) is the LCD because the numerator becomes 1, though 🙂

    Great job!