tag:blogger.com,1999:blog-15628310.post7903404030877820542..comments2021-01-21T13:53:28.761-05:00Comments on Question of the day: What's Wrong With That?Anonymoushttp://www.blogger.com/profile/18153935609499338685noreply@blogger.comBlogger11125tag:blogger.com,1999:blog-15628310.post-6065264313895396732013-11-25T12:29:18.004-05:002013-11-25T12:29:18.004-05:00we can prove that 9.99999...... is a rational numb...we can prove that 9.99999...... is a rational number and all other numbers in this format (repeated form) are rational numbers <br />a = 9.999999...<br />10a = 99.999999...<br />10a - a = 90<br />9a = 90<br />a = 10<br /> by Zafarullah khanAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-15628310.post-9482829374945367042013-11-12T10:31:27.456-05:002013-11-12T10:31:27.456-05:00Abe's response is the correct. The principal d...Abe's response is the correct. The principal difference is the type of number: rationale or non-rationale. You can't mix orange and lemons :):):)<br />When we say 1/3=0.333333..., that is an non-rationale aproximation. The correct is 1/3. The difference is minimum (correct) but there is a difference.<br />So, when you're talk about "aproximation" the equations are correct: "The rationale approximation of the non-rationale 0.999999... is 1"Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-15628310.post-63549283876508977332007-09-15T02:49:00.000-04:002007-09-15T02:49:00.000-04:00sorry, 9.999...=10 :)I am Finnish and we use "," a...sorry, 9.999...=10 :)<BR/>I am Finnish and we use "," as the decimal separatorLeena Helttulahttps://www.blogger.com/profile/07125296730932636809noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-57391685310135683832007-09-15T02:47:00.000-04:002007-09-15T02:47:00.000-04:00It definitely is a trick question!That is one of t...It definitely is a trick question!<BR/>That is one of the extraordinary and magnificient features of real numbers. You can't tell which is the closest smaller number than 10, since 9,999... = 10.Leena Helttulahttps://www.blogger.com/profile/07125296730932636809noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-36663263230173859662007-08-11T10:05:00.000-04:002007-08-11T10:05:00.000-04:00Hello DJ Lower/kkairos, I am the blog author and ...Hello DJ Lower/kkairos, I am the blog author and I believe it is a trick question. I've been wrong enough in the past not to be completely certain about that.Anonymoushttps://www.blogger.com/profile/18153935609499338685noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-84543856742432575732007-08-11T04:44:00.000-04:002007-08-11T04:44:00.000-04:00Yeah. I was a bit worried when I saw Abe's post, b...Yeah. I was a bit worried when I saw Abe's post, but other peoples' took me back to reality and reminded me that, yes, this is the same logic as the 0.999... = 1 proof<BR/><BR/>could the blog author please come forward and let us know whether this was a trick question or whether they just screwed up on this one?Dan Lowerhttps://www.blogger.com/profile/13013884098326991088noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-21711162219825821402007-08-02T11:26:00.000-04:002007-08-02T11:26:00.000-04:00I researched it, and it looks like you're right.9....I researched it, and it looks like you're right.<BR/><BR/>9.99999... is an infinite geometric sum:<BR/><BR/>9*(1/10)^0 + 9*(1/10)^1 + 9*(1/10)^2 = 9 / (1 - 1/10) = 10.<BR/><BR/>I used the definition of geometric sums, a/(1-r). I still say that level of precision argument I pointed out is sort of the crux of the trick in your proof, but apparently 9.99999.. = 10 in our decimal system for some reason. Good post, Mike.Abehttps://www.blogger.com/profile/04424868492071587450noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-47065270466506460832007-08-01T09:15:00.000-04:002007-08-01T09:15:00.000-04:00I'm going to side with anonymous and nick on this ...I'm going to side with anonymous and nick on this one. There's (I believe) nothing wrong with the equations. 9.999... is equal to 10.<BR/><BR/>It was a trick question.Anonymoushttps://www.blogger.com/profile/18153935609499338685noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-68790673182549895152007-08-01T09:11:00.000-04:002007-08-01T09:11:00.000-04:00this can also be displayed in the following patter...this can also be displayed in the following pattern<BR/>1/9= .111111...<BR/>2/9= .222222...<BR/>3/9= .333333...<BR/>...<BR/>8/9= .888888...<BR/>follwoing this logic,<BR/>9/9= .999999... , where the fraction 9/9 also = 1Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-15628310.post-59186579306287966682007-07-31T22:33:00.000-04:002007-07-31T22:33:00.000-04:00I always thought that this was a legitimate proof,...I always thought that this was a legitimate proof, and so the 1 in 89.991 was just a condition of not extending the ...999's.<BR/><BR/>If this proof is invalid, what does that say about the following:<BR/><BR/>1/3 + 2/3 = 1<BR/>1/3 = .33333...<BR/>2/3 = .66666...<BR/>.33333... + .66666... = .99999...<BR/>Therefore .99999... = 1<BR/><BR/>?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-15628310.post-91862095034753545502007-07-31T11:03:00.000-04:002007-07-31T11:03:00.000-04:00The problem comes when you multiply by 10 and subt...The problem comes when you multiply by 10 and subtract. Since 9.999999... is not a rational number (that is, you can't express it in closed form with a fraction, like you could with, say 1.111111... = 10/9), we can only express it to some degree of accuracy (some number of significant digits, that is). Then, when you multiply by 10, the level of precision you chose is moved to the next level. If you do it this way, the problem resolves itself. For example:<BR/><BR/>a = 9.999<BR/>10a = 99.99<BR/>10a - a = 89.991<BR/>9a = 89.991<BR/>a = 9.999<BR/><BR/>The same principle applies for any number of significant figures you choose.Abehttps://www.blogger.com/profile/04424868492071587450noreply@blogger.com