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1+2+4+8+16 is clearly 31

Clearly, it's clearly clear that clear will clearly lose all clear meaning if you use it clearly too much. Clearly

Are we clearly clear?

At 7/21/10 06:05 AM, RubberTrucky wrote:

x-2x=-x=1+2-2+4-4+8-8+...=1.

He added one, but did not subtract by one? 1 is the only digit without an accompanying negative version of itself. Hence, all the other numbers will cancel out, leaving 1 as his answer.

At 7/21/10 11:32 AM, Hoeloe wrote:

At 7/21/10 10:42 AM, Nixot wrote:

At 7/21/10 10:16 AM, Hoeloe wrote: x-2x = infinity - infinity = undefined I believe.

Zero, you twat.

Nope. Infinity is not a conventional number, and cannot be treated as one.

This is indeed correct. Infinity minus infinity can't be 0 or my reasoning is alright. To subtract or add limits you have to make sure the limits are finite first. that's why this deduction won't work.

Let me explain with an example:

There are an infinite number of odd numbers. There are also an infinite number of odd numbers. There are an infinite number of numbers. The number of numbers as a whole is equal to the number of odd numbers plus the number of even numbers, therefore: infinity + infinity = infinity.

Actually, let me pull a math mindfuck here. The number of even numbers and the number of positive integers are the same.
To see this multiply every positive integer with 2. That way I get for every single integer a unique even number. Now take any even number, then half of that number is a positive integer, so we should have accounted for it when we mukltiplied the integers with 2. Because now we have paired integers with even numbers, we can say that the number of even numbers and the number of integers are the same.

If we were to add all the negative numbers notice we can count them as
0,1,-1,2,2,3,3,4,-4,5,-5,...
We can continue like this and you can realise we won't skip any numbers. however by putting them in a sequence, we assign to each positive integer a general integer, so there are equally many general integers as there are positive integers.

Now rational numbers (or fractions). we can regard them as a numerator and a denominator. so essentially 2 integers. Take a lattice with in the vertical direction all integers and in the horizontal direction all integers. So each point in the lattice is a pair of integers. Now count them starting from (0,0) and then the 3x3 square around it then the 5x5 square around that one and so on. Once again, you can see there are equally as many rational numbers than there are positive integers.

Smart Reply: of course, but there are infinitely many of each, because infinity=infinity, there must be the same amount of numbers.

Wrong, Cantor has proven you can't count all the real numbers between 0 and 1 (which are basically decimals but with infinite length and no repetition behind the floating point)
To do this, suppose you can count them, put the first number at the top and then the next number underneat and so on. Take the first numbre behind the floating point of the first element, the second of the second element and so on. Now change every number you've chosen into another one.

If the resultant number is in the sequence it has a position, for instance the 100 th. But because it's the 100 th, the 100th number behind the floating point should be the one you've initially chosen. but you've altered it, so it can't be the 100th number. So you can conclude that the constructed real number is nowhere in the sequence.

Essentially there are far more infinite real numbers then there are integers. So infinity can be larger than infinity.

At 7/21/10 06:12 AM, Shauna wrote: Actually, the answer is lupus.

Actually, the answer is cocks.

At 7/21/10 02:32 PM, RubberTrucky wrote: A lot of good maths.

Yes, this is all true. I was trying to explain a similar concept, but you did it better.