tag:blogger.com,1999:blog-173573593078541140.post3986819396656236289..comments2022-11-14T10:59:25.838+00:00Comments on rebellionkid's blog: RESPOST Why I love maths.Rebellionkidhttp://www.blogger.com/profile/05285549817197747799noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-173573593078541140.post-10898167736694269972009-11-27T01:09:20.639+00:002009-11-27T01:09:20.639+00:00ThreeWords: I'm not sure quite what you mean s...ThreeWords: I'm not sure quite what you mean sorry, basically what I tried to show is that whatever pairing up you do at all there will always be one missing, and even if you put that one back in you will be able the use the same argument, no matter what pairing you imagine there will always be something left out. Sorry, if I've misunderstood your point. Try again if you think I've made a mistake, I may get it the second time.<br /><br /><br /><br />Anonymous: Your proof is nice and simple, but it does lead to a problem, that you may get people saying "well of course you cant pair up all the elements if you do it like that, but there might be some other way of doing it". It may be that there's a way to answer this objection, but at half midnight i cant work it out lol, but your method does have the advantage of being simple to understand.<br /><br /> I think I may have failed to explain an important point properly that may be the cause of your objection at the start though. You say that this extra goat I've found can always be paired up by finding another pot. The point I should have stressed is that the way I found the extra goat works for absolutely any pairing you may care to imagine.<br /><br /> So suppose we add this extra goat into the line up, then the pairing we have imagined is still missing one (that I haven't taken out by the way, so it wasn't there in the first pairing either) so it's still the case that there are more goats. I should have explained more clearly that adding the extra goat doesn't get you out of the mess. Hope that clears things up, if not I'll try again.<br /><br /> In fact, I'll add in the second beautiful wonderful thing (tm). This might help explain the difficulty of your proof:<br /><br />Theorem: The size of the set of rational numbers (for people who dont like technical terms, i mean fractions with whole numbers top and bottom) is the same as the size of the set of whole numbers.<br /><br />Proof: I can set up a pairing, (which I cannot do for the real numbers with infinite decimal expansions) which goes like this: <br /><br />1: 1/1<br />2: 1/2<br />3: 2/1<br />4: 1/3<br />5: 2/2<br />6: 3/1<br />7: 1/4<br />8: 2/3<br />9: 3/2<br />etc<br /><br />If it's not clear how I got that, picture a grid with the numerator from 1 to infinity on the bottom axis, and the denominator from 1 to infinity on the vertical axis, then take diagonal slices like:<br /><br />.\.\.\.\.\.<br />.\.\.\.\.\. (sorry for the rubbish<br />.\.\.\.\.\. drawing, it's hard to<br />.\.\.\.\.\. do in ascii)<br /><br />and work up the diagonals in order. Every possible pair of numbers will eventually be reached this way, no line is infinitely long, so how ever large the fraction it matches to some number. So a pairing exists, which I said was our measure of the size of the set of things. <br /><br />Why cant I do something like this for all the real numbers? Because pi and root two and an infinite (of the bigger kind) amount of numbers like them can never be written as a fraction and so will never be reached by this.<br /><br /><br /><br /><br /><br />All three of you, if anything has confused you or you think I've made a mistake (I probably have it's very late) please comment again. Thanks for your responses, it was nice to get a response for once lol.Rebellionkidhttps://www.blogger.com/profile/05285549817197747799noreply@blogger.comtag:blogger.com,1999:blog-173573593078541140.post-81836664538507023842009-11-26T23:42:10.265+00:002009-11-26T23:42:10.265+00:00AH. I see what you did. You are confusing the dens...AH. I see what you did. You are confusing the density of an infite set and the length. <br /><br />See, both sets are the exact same length, but between 2 intergers exist an infite amount of real numbers. <br /><br />Yet you can pair every real number with an interger. Take the next number up from an interger and pair that with 2, then the next with three and so forth. <br /><br />This is similar to fitting two infinite sets within a single one, by alternating the placement. <br /><br />However, infinity has some weirder properties such as that adding anything to it results in itself. <br /><br />If you want amazing math, do calculus with natural logs and the relationship with all other powers of constants.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-173573593078541140.post-69073417292300644772009-11-26T19:41:03.113+00:002009-11-26T19:41:03.113+00:00I would agree with the statement that there are di...I would agree with the statement that there are different sized infinities, but I don't agree with the proof. <br /><br />Although by the method you specified you can always come up with a new number goat set, you can also always come up with a number in the pot set to match it. Also if you take the infinite set of all whole numbers and add "0." to the front of all its elements, you will have created the second set. By this method it could be argued that these sets are in fact the same size.<br /><br />However I think that the original statement is true and will try to create an example that shows it(I'm just doing this quickly so I might mess up). Consider the set of all non-negative integers(let this be the pot set) and the set of all non-negative real numbers(let this be the goat set). <br /><br />All the integers are real numbers, however not all the real numbers are integers. That is the integers are a subset of the real numbers, but the real numbers are not a subset of the integers.<br /><br />Now for every element of the integers you associated with its equivalent number in the real numbers (1 with 1, 243124 with 243124, etc.). This will cause all elements of the integers set to be paired with an element of the real numbers set. However, numbers which are not integers in the real numbers set(0.5, pi, 12.1, etc.) have no pair in the integers set and there are no elements available to pair with. You have goats left over.<br /><br />Therefore it can be said that the set of all non-negative real numbers is "bigger" than the set of all non-negative integers, despite the fact that they are both infinite.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-173573593078541140.post-68983543664837685352009-11-26T19:10:51.329+00:002009-11-26T19:10:51.329+00:00I don't want to be mean/stupid, but surely tha...I don't want to be mean/stupid, but surely that doesn't work?<br /><br />Your theory works, but only if you assume that an a decimal will be left over. And since both the integers and decimals are infinite, there should be none left over.<br /><br />All you've done is say "There's two infinite sets, but one has more in. Therefore, there are two kinds of infinity." It is only true because you set up the senario that wayThreeWordshttp://www.escapistmagazine.com/profiles/viewnoreply@blogger.com