Puzzle Description:

This proves that 1 equals .9999 repeated. I was in awe when I saw this proof in high school. I'm glad I found some math geeks to share this with. ;-)

#1: Gator (Gator) on Sep 22, 2010 [HINT]

By looking at the picture I can guess how to finish this one, but I cannot see a way to logically figure out the last part of the puzzle. Am I missing something?

I did it by logic alone. Just had to be very careful in last 5% or so. :o)

I'm having trouble with the solution at the very end as well. I tried internal edge logic and two-way logic, but no go.

Mrs. Thing (or Ron), could you share how you did it?

Ron, did your high school math teacher put one over on you? We either round both 10x and 9x or we don't round either. If x = .9999 repeated, then 9x = 8.9999 repeated, not 9. Then x still equals .9999, and 1 does not equal .9999. What am I missing?

The hard part to grasp is that the repeating doesn't end. It continues indefinitely. This is one of the core concepts of what is called limits in math. .9 repeated can also be represented as an infinite geometric series. Basically, 9/10 + 9/100 + 9/1000 + ... with a starting value of 9/10 and a ratio of 1/10. The formula for calculating the value of an infinite geometric series is:

S = a1 / (1 - r) where S is the sum, a1 is the first term of the series, and r is the ratio. Thus:

S = (9/10) / (1 - 1/10) = (9/10)/(9/10) = 1

There are many other ways that this can be proven also with Ron's as one of the ways.

Got it. Thanks, Gator. I didn't understand Ron's proof but I do understand your explanation of an infinite geometric series and the formula for calculating it. Now I can email my friend the math PhD and retired university math professor and show him how smart I am. I may not mention you or Ron, lol.

By the way, that was fun. Thanks, both of you.

I do remember being taught this and it didn't make any more sense than it does now. It was a fun puzzle though.

Goto next topic

You must register and log in to be able to participate in this discussion.