# are there more multiples of 3 or 7?

31 posts

 my friend and I talking about this, he say equal cus both infinite, i say: lim x→ inf (floor(x/3) > floor(x/7)) Originally posted by UnknownGuardian: edited, http://www.wolframalpha.com/input/?i=lim+x+to+infinity+%28floor%28x%2F3%29+%3E+floor%28x%2F7%29%29 EDIT x1 for your edit: infinity x 2 might be less than infinity EDIT x2 I think you have to separate limits, not group em. Both those limits are infinity. Not sure what you were going to prove with it either. Considering that we proved that there are the same number of rational numbers as integers your friend is correct. Both sets are the same size. Multiples of three can be enumerated as follows: f(x) = 3x, where x is an integer. Multiples of seven can be enumerated similarly: g(x) = 7x, where x is an integer. Both of these functions are one-to-one mappings, and what’s more, the two can be combined: h(x) = 3x/7, where x is a multiple of 7. h(x) provides a one-to-one mapping from the set of integer multiples of 7 onto the set of integer multiples of 3. In other words, every multiple of 7 can be paired with a multiple of 3, and no numbers from either set will be left out. Therefore, the sets are the same size. Q.E.D. See also: Hilbert’s Grand Hotel. Originally posted by player_03:Multiples of three can be enumerated as follows: f(x) = 3x, where x is an integer. Multiples of seven can be enumerated similarly: g(x) = 7x, where x is an integer. Both of these functions are one-to-one mappings, and what’s more, the two can be combined: h(x) = 3x/7, where x is a multiple of 7. h(x) provides a one-to-one mapping from the set of integer multiples of 7 onto the set of integer multiples of 3. In other words, every multiple of 7 can be paired with a multiple of 3, and no numbers from either set will be left out. Therefore, the sets are the same size. Q.E.D. See also: Hilbert’s Grand Hotel. fascinating :) Worth mentioning though that your friend is correct but for the wrong reason: “equal cus both infinite” is wrong. For instance, there are infinitely many integers, and infinitely many real numbers, but there are infinitely more real numbers than integers. The two sets are not in the same class of infinity, and you can prove it. As for the limit when x tends to infinity of floor(x/3) / floor(x/7), it’s 7/3. I liked this one: http://www.youtube.com/watch?v=VjZyOTES6iQ Oh, and it gets a bit less physics related from 3:18 What’s `infinity / infinity`? 1. Originally posted by SWATLLAMA:1. False. It’s undefined. infinity / infinity = 1 → (infinity + infinity) / infinity = 1 → (infinity / infinity) + (infinity / infinity) = 1 1+1=1 ERROR http://www.philforhumanity.com/Infinity_Divided_by_Infinity.html Originally posted by ErlendHL:What’s `infinity / infinity`? This is one of the cases where we can’t really say anything. Other such cases include: 0*inf, 1^inf, inf – inf Originally posted by someone93:Originally posted by ErlendHL:What’s `infinity / infinity`? This is one of the cases where we can’t really say anything. Other such cases include: 0*inf, 1^inf, inf – inf can you explain why 1 ^ infinity is obscure? o-o I can’t remember specifics. But we showed it in my Uni. intro math. Originally posted by someone93:I can’t remember specifics. But we showed it in my Uni. intro math. Originally posted by qwerberism:Originally posted by someone93:Originally posted by ErlendHL:What’s `infinity / infinity`? This is one of the cases where we can’t really say anything. Other such cases include: 0*inf, 1^inf, inf – inf can you explain why 1 ^ infinity is obscure? o-o I think it’s similar to why 0*inf is obscure. 0 is similar to 1/infinity. so 0*inf is similar to infinity/infinity, which is obscure. 1^inf = (1+0)^inf = (1-0)^inf which is similar to (1-1/infinity)^infinity and (1+1/infinity)^infinity, which are obscure. Basically if you were to analyze infinity as some number x approaching infinity, then 1 is really some number y approaching 1. Then y^x isn’t necessarily 1 and is obscure. Originally posted by qwerberism:Originally posted by someone93:Originally posted by ErlendHL:What’s `infinity / infinity`? This is one of the cases where we can’t really say anything. Other such cases include: 0*inf, 1^inf, inf – inf can you explain why 1 ^ infinity is obscure? o-o Let’s say 1^inf = x ln(1^inf) = ln(x) inf*ln(1) = ln(x) inf*0 = ln(x) `1 ^ infinity` should `= 1` I think?? ```3/0 = inf inf * 0 = 3 ``` 100% legit. ```Proof: log b to the base a = (log b)/(log a) Example: log 8 base 2 = 3 ---- 2^3 = 8 (log 8)/(log 2) = 3 log 8 base 1 = (log 8)/(log 1) = (log 8)/0 = infinity It follows that 1^infinity = 8 Similarly log 9 base 1 = infinity It follows that 1^infinity = 9 1^infinity can have any positive value greater than or equal to 1. 1^infinity >= 1 1^(-infinity) = 1/1^(infinity) 0 <= 1^(-infinity) <= 1 1^infinity is indeterminate (undefined).``` x/0 is not infinity, it’s indeterminate and contextual. Originally posted by BigJM:x/0 is not infinity, it’s indeterminate and contextual. x/0 = a (or a*0 = x), for x != 0, is undefined, not indeterminate. There is not a single solution. 0/0 = a (or a*0 = 0) on the other hand is indeterminate. There is not a unique solution (it’s true for every value of a). It is unable to be determined; therefore, it is indeterminate. Originally posted by BigJM:It is unable to be determined; therefore, it is indeterminate. You would be correct. An indeterminate system is a system of simultaneous equations (especially linear equations) which has infinitely many solutions or no solutions at all. The system may be said to be underspecified. I think it’s assuming that x is in the extended complex plane. That’s why I said it’s meaning is contextual. Originally posted by BigJM:I think it’s assuming that x is in the extended complex plane. That’s why I said it’s meaning is contextual. Oddly I got the same result when I try `Re[x]/0` Ran it through Mathematica 8: ```In[1]:= Re[x]/0 During evaluation of In[1]:= Power::infy: Infinite expression 1/0 encountered. >> Out[1]= ComplexInfinity ```