BMAD 65 Posted October 19, 2013 Report Share Posted October 19, 2013 Here is a classic exercise: Which of the following are true and why? 0.99999..... < 1 0.99999..... = 1 0.99999..... > 1 Quote Link to post Share on other sites

0 Solution ThunderCloud 5 Posted October 20, 2013 Solution Report Share Posted October 20, 2013 Sum_{i = 1 to infinity](0.1^i) = 1/9 as a geometric series (also easily proven algebraically). 0.999... = 9 * (0.111...) = 9 * Sum_{i = 1 to infinity](0.1^i) = 9 * 1/9 = 1. The equality is therefore true. By extension, both of the inequalities are not true. Quote Link to post Share on other sites

0 bonanova 85 Posted October 20, 2013 Report Share Posted October 20, 2013 The easiest way to choose among the three options -- one that does not require the invocation of limits, Cauchy sequences, Dedekind cuts or delving into the theory of real numbers -- is to note the decimal representation of the rational number 1/3. 1/3 = 0.3333 ... Multiply by 3. Quote Link to post Share on other sites

0 BMAD 65 Posted October 20, 2013 Author Report Share Posted October 20, 2013 So are we saying only one case is true? Quote Link to post Share on other sites

0 ThunderCloud 5 Posted October 20, 2013 Report Share Posted October 20, 2013 So are we saying only one case is true? Yes. Both sides of the relations specify real numbers, which are well-ordered. So, exactly one of the relations must be true. Quote Link to post Share on other sites

0 phil1882 13 Posted November 4, 2013 Report Share Posted November 4, 2013 hmm. there's a mathematician on youtube, norm wildberger, whom i kinda like who would disagree. the problem being that we cant actually "know" infinity. try writing infinite 9's. you can't. so at best we can say 0.9999999 or wherever you care to stop approximately equals 1. to prove this point, accurately calculate pi +e +sqrt(2). if you're being honest, you would calculate it as pi + e +sqrt(2). Quote Link to post Share on other sites

0 Prime 15 Posted November 4, 2013 Report Share Posted November 4, 2013 When I was in 5th grade, my math teacher gave the class the following proof: (10x - x)/9 = x 10*0.999... = 9.999... _ 9.999... 0.999... 9 9/9 = 1 QED Quote Link to post Share on other sites

0 bonanova 85 Posted November 5, 2013 Report Share Posted November 5, 2013 hmm. there's a mathematician on youtube, norm wildberger, whom i kinda like who would disagree. the problem being that we cant actually "know" infinity. try writing infinite 9's. you can't. so at best we can say 0.9999999 or wherever you care to stop approximately equals 1. to prove this point, accurately calculate pi +e +sqrt(2). if you're being honest, you would calculate it as pi + e +sqrt(2). The notation convention 0.999 ... denotes an endless string of 9's. You can't write the decimal representation of 1/3, but you can write 0.333 ..., Thus, given the meaning of that notation, one can meaningfully write 1/3 = 0.333 ... Similarly one can meaningfully write 1 = 0.999 ... Quote Link to post Share on other sites

0 phil1882 13 Posted November 5, 2013 Report Share Posted November 5, 2013 well, not entirely sure i agree. for example, how would you denote 1/13? you need a consistent way of denoting any rational number, not just ones that repeat the last digit. personally, i like the bar method, where the repeated values have a bar over the top over the length of the repeating part. so _ 0.9 = 1 would be more accurate (although more cumbersome.) Quote Link to post Share on other sites

0 phil1882 13 Posted November 5, 2013 Report Share Posted November 5, 2013 here's my challenge for you. accurately add 1/13 to 45/89 in decimal representation. Quote Link to post Share on other sites

0 ThunderCloud 5 Posted November 5, 2013 Report Share Posted November 5, 2013 well, not entirely sure i agree. for example, how would you denote 1/13? you need a consistent way of denoting any rational number, not just ones that repeat the last digit. personally, i like the bar method, where the repeated values have a bar over the top over the length of the repeating part. so _ 0.9 = 1 would be more accurate (although more cumbersome.) ______ 1/13 = 0.076923 The bar covers the part that repeats. here's my challenge for you. accurately add 1/13 to 45/89 in decimal representation. This can be done, but would be rather tedious (particularly as the decimal expansion for 45/89 has a relatively long repeating digit string). The sum will also have a repeating pattern, of some size less than or equal to the LCM of the repeating digit string lengths of the two operands. Quote Link to post Share on other sites

0 ThunderCloud 5 Posted November 5, 2013 Report Share Posted November 5, 2013 well, not entirely sure i agree. for example, how would you denote 1/13? you need a consistent way of denoting any rational number, not just ones that repeat the last digit. personally, i like the bar method, where the repeated values have a bar over the top over the length of the repeating part. so _ 0.9 = 1 would be more accurate (although more cumbersome.) I might have missed the gist of this question the first time around... Another way to express 1/13 is 0.076923076923076923... It is clumsy compared with the bar method in this example (unlike .999...), but still understandable. Quote Link to post Share on other sites

## Question

## BMAD 65

Here is a classic exercise: Which of the following are true and why?

0.99999..... < 1

0.99999..... = 1

0.99999..... > 1

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## 11 answers to this question

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