tag:blogger.com,1999:blog-9182705499898252496.post8079147349585283338..comments2019-01-22T13:40:10.758-05:00Comments on Bill the Lizard: Six Visual ProofsBill the Lizardhttp://www.blogger.com/profile/09810099093752485841noreply@blogger.comBlogger72125tag:blogger.com,1999:blog-9182705499898252496.post-91580746739755319222018-09-16T16:05:30.283-04:002018-09-16T16:05:30.283-04:00When it comes to weight loss. Only a few supplimen...When it comes to weight loss. Only a few suppliments can get the job done without any side effects. The best one being Rapid Tone Diet. Watch a review here: <a href="http://www.youtube.com/watch?v=Zu77ZXLW2Bg" rel="nofollow">Rapid Tone Diet.</a><br />Alan Bakerhttps://www.blogger.com/profile/17731654295537657694noreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-91516402396053666092018-09-02T08:25:39.870-04:002018-09-02T08:25:39.870-04:00I love Premier Diet Keto. It is the best fat loss ...<a href="http://www.youtube.com/watch?v=4ge7qSzACdo" rel="nofollow">I love Premier Diet Keto. It is the best fat loss supplement that I've ever tried with zero side effects. If you're skeptical or have any doubt about what I'm saying watch a video review and find it out yourself. Click here Premier Diet Keto </a><br />Alan Bakerhttps://www.blogger.com/profile/18241960961537916241noreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-91172352854992189952018-08-24T10:50:02.577-04:002018-08-24T10:50:02.577-04:00If you want a good fat burning supplement then Pre...If you want a good fat burning supplement then Premier Diet Keto is the one for you. Using ingredients like Garcinia Cambogia and Forskolin its both effective and free from side effects. For review click <a href="http://www.youtube.com/watch?v=4ge7qSzACdo" rel="nofollow"> here </a><br />Alan Bakerhttps://www.blogger.com/profile/17731654295537657694noreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-12365138975527547262014-07-19T23:50:51.391-04:002014-07-19T23:50:51.391-04:00I couldn't see #6 until I saw the image of nde...I couldn't see #6 until I saw the image of ndef. So that color shading would indeed help.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-32499690520424682502011-04-16T21:53:43.257-04:002011-04-16T21:53:43.257-04:00Really, really nice post!Really, really nice post!Maximenoreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-25467088029572536112010-06-18T23:49:45.714-04:002010-06-18T23:49:45.714-04:00The 1/2 + 1/4 + 1/8 + 1/16 + .. = the sum of 1/2^n...The 1/2 + 1/4 + 1/8 + 1/16 + .. = the sum of 1/2^n - 1, where n ranges from 0 to infinity. By basic analysis, this is simply a geometric series, sum of r^n, where r = 1/2, therefore it is convergent, and the answer is simply derived either by knowing the geometric series theorem or by using taylor series expansion around 0, to get 1/(1-r) - 1 ( remember we had to subtract one for the series to work out right.) therefore you arrive at 2 - 1 = 1, the desired answer.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-47128259125060660312010-06-12T14:03:22.767-04:002010-06-12T14:03:22.767-04:00See Alex's work, as a 3rd grader, showing 1/3+...See Alex's work, as a 3rd grader, showing 1/3+1/9+1/27...->1/2 Below his work on 1/2+1/4+... at http://www.mathman.biz/html/alexs.html. He said 1/3 + 1/2of1/3 = 1/2 which I didn't see until I looked carefully at his picture.<br />Don Cohen-The MathmanDonald Cohenhttp://www.mathman.biz/noreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-84711926737125500532010-06-12T06:08:27.096-04:002010-06-12T06:08:27.096-04:00more evidence that the universe is an integer trut...more evidence that the universe is an integer truth value of existence that's dividing over time into 'smallest discernably discreet parts' of universe...<br /><br />...and all the 'smallest discernably discreet parts' add up to make-and-occupy-positions-within the qcd lattice... all derivable from number theory...<br /><br />why not?<br /><br />:PAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-8792571960707640642010-05-29T09:03:04.763-04:002010-05-29T09:03:04.763-04:00Josh Koerner,
I love the golf ball photo you used....Josh Koerner,<br />I <i>love</i> the golf ball photo you used. You can really make an abstract concept more accessible by 1) showing a picture of it, and 2) putting it in terms of a real-world problem.<br /><br />As for blogging advice, a lot of people will tell you to write what you know. This is fine advice, but I tend to take a slightly different approach. I like to write about things as I'm learning them. If I can explain a concept good enough to put it in writing, I feel like I've come to a pretty good understanding of it.<br /><br />I hope that's helpful, and good luck with your blog.Bill the Lizardhttps://www.blogger.com/profile/09810099093752485841noreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-16570562522693265312010-05-27T23:36:36.838-04:002010-05-27T23:36:36.838-04:00Thank you! That was extremely helpful. I am new to...Thank you! That was extremely helpful. I am new to blogging but find it exciting and useful. I recently uploaded your proof to <a href="http://iceaccis.wordpress.com/2010/05/26/counting/" rel="nofollow">my blog</a> and used it to help explain the problem.<br /><br /><br />I was wondering if you would take a look and possibly give me some pointers about a successful blog. Your's was easy to use and informational. I would like to improve mine and would appreciate any suggestions. Thank you in advance for your time.<br /><br /><br /><br />Thank You,<br /><br />Josh KoernerJosh Koernerhttps://www.blogger.com/profile/01239931677692935246noreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-65852599190109669772010-05-25T19:53:07.857-04:002010-05-25T19:53:07.857-04:00Hi Josh, thanks for commenting. There's an ol...Hi Josh, thanks for commenting. There's an old story about Gauss that might shed some light on the fourth proof. (I'm quoting from <a href="http://www.billthelizard.com/2009/01/gauss.html" rel="nofollow">my own post about Gauss</a>, but this story has been retold hundreds of times.)<br /><br />"When Gauss was in school at age 10, his teacher, perhaps needing a quiet half hour, gave his class the problem of adding all the integers from 1 to 100. Gauss immediately wrote down the answer on his slate (this would have been in the year 1787). He had quickly noticed that the sum (1 + 2 + 3 + ... + 100) could be rearranged to form the pairs (1 + 100) + (2 + 99) + (3 + 98) + ... + (50 + 51), and that there were 50 pairs each equaling 101. This reduced the problem to the simple product 50 x 101 = 5050."<br /><br />The fourth proof here is illustrating the same idea. It's just arranging the numbers in a different way than Gauss did over 200 years ago.<br /><br />I hope that helps.Bill the Lizardhttps://www.blogger.com/profile/09810099093752485841noreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-77181583502635837512010-05-25T15:42:55.619-04:002010-05-25T15:42:55.619-04:00After looking at the fourth proof I set up a displ...After looking at the fourth proof I set up a display of a triangle of golf balls. I wanted to know if this proof stays true when you increase the numbers. Even with replete rows and columns the proof stayed true. I was hoping you could further help my understanding as to the reasons behind the logic of the fourth proof.Josh Koernerhttps://www.blogger.com/profile/01239931677692935246noreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-36968077809517439332010-05-15T01:09:13.683-04:002010-05-15T01:09:13.683-04:00Can you visualize a proof of the fact that
sum of ...Can you visualize a proof of the fact that<br />sum of reciprocals of squares of the positive integers equals pi squared by 6?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-86178901589570583032010-03-31T19:19:25.433-04:002010-03-31T19:19:25.433-04:00The 6th visual proof is simpler. Einstein used a s...The 6th visual proof is simpler. Einstein used a single line from the hypotenuse making two other, enclosed, right triangles. The area of each of the three triangles (two inner, one outer) is proportional to the square of its hypotenuse. It is the same proportion for all three similar triangles.<br /><br />Visually it can then be seen that a^2+b^2=c^2.<br /><br />This visual proof would require the "visualization" of the equation <br /> kA1^2+kA2^2=kA3^2<br />where A is an area, and k the proportional constant.<br /><br />This comes from the author Manfred Schroeder ("Fractals, Chaos, Power Laws"), who got it from Schneior who got if from Ernst Strauss, who got it from Albert who came up with it during a home schooling session with his Uncle Jacob.Jameshttps://www.blogger.com/profile/08625273817903506842noreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-39521819771930875732010-03-30T06:27:34.765-04:002010-03-30T06:27:34.765-04:00Ah excellent, I get it now - both ways! Cheers MaJ...Ah excellent, I get it now - both ways! Cheers MaJJ and Bill the Lizard.Bob T. Buildernoreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-11655974689921810642010-03-29T12:36:23.885-04:002010-03-29T12:36:23.885-04:00Bob T. Builder,
Just to add to MaJJ's explanat...Bob T. Builder,<br />Just to add to MaJJ's explanation, there are exactly 3 of each size of the smaller squares. One of each size is shaded green, so 1/3 of the whole figure is green, and the remaining 2/3 is grey.<br /><br />Another way of looking at it is if you focus on the green squares, look to the right of each one. The grey square to the right of each green square is the same size. Now look directly up from each green square and you'll notice the same thing. So the area to the right of the green squares and the area above the green squares are each equal to the green area.Bill the Lizardhttps://www.blogger.com/profile/09810099093752485841noreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-51899136439897833162010-03-29T11:08:29.948-04:002010-03-29T11:08:29.948-04:00Bob: When looking at the first iteration (four squ...Bob: When looking at the first iteration (four squares, each with area 1/4), try to ignore the right-top one. It will be dealt with in the second iteration. Now you see that the green square takes 1/3 of the three squares. And it's like that with all the other iterations.<br /><br />(I hope I wrote it understandably :) )MaJJhttps://www.blogger.com/profile/09326001140797748108noreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-85368675924504456702010-03-29T07:46:24.320-04:002010-03-29T07:46:24.320-04:00Can anyone shed some light on the first one? I jus...Can anyone shed some light on the first one? I just don't see how you can see 1/3 in the green (or 2/3 in the grey)?Bob T. Buildernoreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-10037525244473857822009-12-23T23:52:16.530-05:002009-12-23T23:52:16.530-05:00Anonymous,
No, you're wrong about that. 0.999...Anonymous,<br /><br />No, you're wrong about that. 0.999... is definitely equal to 1. This has been accepted by professional mathematicians for a very long time, and is taught in text books. There are many proofs of it <a href="http://en.wikipedia.org/wiki/0.999...#Proofs" rel="nofollow">here</a>.Bill the Lizardhttps://www.blogger.com/profile/09810099093752485841noreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-40779266130963344832009-12-23T23:17:51.036-05:002009-12-23T23:17:51.036-05:00while they are interesting, i must say they are al...while they are interesting, i must say they are all wrong. these all work off the assumption that 0.99999... means that the nines go on forever. they never end and they are "infinite".<br /><br />however you cannot say that 0.99999... = 1, because there is no way that something infinite(0.999...) can equal something finite. one (1) is finite. we accept it as a real number, a rational number, and a whole number. the number one has a definite end and therefore cannot equal something that has no endAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-4505508230742338242009-08-04T13:05:32.714-04:002009-08-04T13:05:32.714-04:00ePlanner,
I see your point about #6. I wonder if ...ePlanner,<br />I see your point about #6. I wonder if I should have labeled the a and b squares in the bottom left figure? At any rate, I started with the top figure and copied/pasted elements from it to construct the bottom two, so I can promise that the triangles are all equal. :)<br /><br />On #5, it did seem to be that fact that n was <i>not</i> the last number in the sequence that confused a few people. Moving the calculations over to the RHS does seem to improve the clarity. Thanks for pointing this out. I'll keep it in mind next time I create a similar post.Bill the Lizardhttps://www.blogger.com/profile/09810099093752485841noreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-9243290352018741402009-08-04T12:16:33.034-04:002009-08-04T12:16:33.034-04:00#6) Yeah true. I just think the tile one leaves no...#6) Yeah true. I just think the tile one leaves no doubt as to the size of each area, where as one could believe that the grey triangles in both the bottom images are different in size).<br /><br />With #5, the formula is of course 100% correct, but I believe the confusion comes from people having to do a calculation in the heads to figure out the value of n (i.e. n = (final number + 1)/2). Instead, they are mistaking n as being the final number (as it is usually represented as such).<br /><br />This image http://img217.imageshack.us/i/proofx.png/ uses a formula that has this step worked out. I also added a few rows at the bottom for further clarity, but these are not required.<br /><br />Cheers for the interesting topic, been awhile since I have looked at this sort of stuff and #5 even confused me for a bit.ePlannerhttps://www.blogger.com/profile/13197296275438591935noreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-11399566000473053092009-08-04T11:29:32.291-04:002009-08-04T11:29:32.291-04:00ePlanner,
#6 proves the Pythagorean theorem by sho...ePlanner,<br />#6 proves the Pythagorean theorem by showing that the square of side c (the bigger, dark green square) is equal in area to the sum of the two smaller, lighter green squares. It does this by inscribing them in squares of equal size, and showing that the area left over (the four gray triangles) are equal.<br /><br />There are literally hundreds of published proofs of the Pythagorean theorem, but this is my favorite due to its simplicity.<br /><br />As for #5, I was hoping that the numbers at the bottom would help by associating the number of dots in each channel with the numbers 1 + 3 + 5 = ... in the formula. Maybe I should have flipped the image so those numbers were at the top?Bill the Lizardhttps://www.blogger.com/profile/09810099093752485841noreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-55027980717277178142009-08-04T10:10:34.738-04:002009-08-04T10:10:34.738-04:00Very nice.
Proofs 1-5 are accurate. Anyone disagr...Very nice.<br /><br />Proofs 1-5 are accurate. Anyone disagreeing is just not misunderstanding something. <br /><br />I do not see how 6 displays the proof however, but have seen another example using tiles), with a=3x3 (9 squares), b=4x4 (16 squares) and c=5x5 (25 squares).<br /><br />I love how proof 1 clearly shows that the green area is 1/3 from the top left, bottom left and bottom right squares, with the same being true for the one left over and so on.<br /><br />The confusion which some people are having with number 5 is that they are confusing the number (from the X axis of the graph) with n, where n is in fact the number minus 1 and divided by two.ePlannerhttps://www.blogger.com/profile/13197296275438591935noreply@blogger.comtag:blogger.com,1999:blog-9182705499898252496.post-12110355982095490442009-08-02T09:07:18.894-04:002009-08-02T09:07:18.894-04:00I love these visual models. About 10-12 years ago,...I love these visual models. About 10-12 years ago, when I was homeschooling my youngest. We had wonderful worksheets that were similar to the 3rd example. It was a great way to really drive home how fractions work.<br /><br />How pleased am I that she is now pursuing math in grad school!Marshahttps://www.blogger.com/profile/14530890819451603472noreply@blogger.com