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04-29-2010, 04:46 AM
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No but really: Squaring a 2 digit number ending in the number 5. 65*65 1. Multiply the first digit one more than itself. (in this example one more than 6 is 7) 6*7 = 42 2. Put 25 after the answer. The answer will always end in 25. 65*65 = 4225 How is this done? Squaring a binomial. It only works however when the number you are squaring ends in 5. 1 = ............................................1= 1*1 1+2+1 =.....................................4 =2*2 1+2+3+2+1=..............................9 =3*3 1+2+3+4+3+2+1=..................... 16=4*4 1+2+3+4+5+4+3+2+1=..............25=5*5 and so on I like math parlor tricks. |
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04-29-2010, 04:58 AM
Exactly. I feel that it really does not matter because china will not ask for their money anytime soon since it would just end up in us calling in debts which no one really has or wants to pay back.
But damn I can not believe .9999999 = 1 can be proven with a geometric sequence. |
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04-29-2010, 08:53 AM
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But if you think about it a bit, it does seem intuitively reasonable that 0.999... = 1. Since you can approximate 0.999... as close to 1 as you like simply by listing enough 9s after the decimal, there's no reason to suspect that the infinitely repeating decimal is not the same as 1. And here's another way to think about it, if you remember that 1/3 has the infinitely repeating decimal 0.333.... Well, obviously 3 * 1/3 must equal 1. But if you take the repeating decimal representation of 1/3 and multiply that by 3, you would get 0.999.... Therefore 0.999.... must be equal to 1. I'm sure the geometric sequence is a much more rigorous way to demonstrate the equality, though. |
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04-29-2010, 08:53 PM
Good point with the ... I could not find a way to do a repeating decimal.
Also If you think about it 1 minus .99999.... people may try and say .000000.. with a 1 somewhere on the end but the act of placeing the 1 ends it and then it is not repeating forever. but if you look at t as an additive like .9.../10^1 +.9.../10^2 + .9..?10^3.... then you see a pattern of .9.../10^k where the pattern begins at k=1 and continues indefinitely then you can represent it as .9.../10^2/.9.../10^1 which equals 1/10 and if you plug that r value into the formula .9.../1-r you get .9.../.9... or 1. |
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05-12-2010, 12:44 PM
You can easily define a math where the infinitesimal exists, and consequently show that .999... doesn't equal 1. Proof of nonexistence is only valid within the relevant universe of discourse, and it is unfortunate that you went through a proof without defining its limitations.
My own opinion on the matter is that if your math contains pi, 0.999... doesn't equal 1. |
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