Proving 0.99999... = 1: A Wikipedia Insight

[nazgjunk]

0.99999... = 1?

I found the following proof on Wikipedia. It looks fine, but I can hardly imagine it to be right...

Another kind of proof adapts to any repeating decimal. When a fraction in decimal notation is multiplied by 10, the digits do not change but the decimal separator moves one place to the right. Thus 10 × 0.9999… equals 9.9999…, which is 9 more than the original number. Subtracting the smaller number from the larger can proceed digit by digit; the result is 9 − 9, which is 0, in each of the digits after the decimal separator. But trailing zeros do not change a number, so the difference is exactly 9. The final step uses algebra. Let the decimal number in question, 0.9999…, be called c. Then 10c − c = 9. This is the same as 9c = 9. Dividing both sides by 9 completes the proof: c = 1.

You can find the full article http://en.wikipedia.org/wiki/Proof_that_0.999..._equals_1" .

 

[HallsofIvy]

"I do not understand it, therefore it must be wrong!"

It assumes some technical facts but it is, in fact, completely valid. Why do you say "I can hardly imagine it to be right..."??

Would you have a problem with this: if x= 0.3333... then 10x= 3.3333333... so, subtracting, 9x= 3 and x= 1/3. Would it surprise you to learn that 1/3= 0.3333...? If not why does it bother you to learn that
3/3= 0.99999...?

This thread has been locked since it has all been said before. Look in the archives:
https://www.physicsforums.com/showthread.php?t=5513