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B Infinity times zero

  1. Jan 25, 2017 #1
    What is infinity times zero? Isn't it zero? I mean, infinity times zero is the same as zero an infinite amount of times, and adding zero infinitely would give zero because even though zero is always being added, this process is the same as nothing ever being added to zero.
     
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  3. Jan 26, 2017 #2

    Student100

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    Simple rules of arithmetic don't apply to infinity, and as such ## \infty \times 0## doesn't mean anything. It's indeterminate.

    That leads to all kinds of confusing arguments, what then is ##.0000000000000000000001 \times \infty##? What about ##
    \lim_{x \to \infty} (x \times 1/x)##, why is that one in this case?
     
  4. Jan 26, 2017 #3
    Adding something that is finite an infinite amount of times would not lead to a finite value, it would lead to an infinite value, so maybe that is the answer to the your first question.

    The second question is about limits (which I could use help with), but x/x = 1, so maybe there is no need to consider the variable x as it goes to infinity, so there is no need to consider a limit.
     
  5. Jan 26, 2017 #4

    Student100

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    You're still thinking about infinity as an actual amount, it doesn't belong to any of the numbers in which arithmetic depend on. It's a concept. What infinity are we multiplying it by? Is there a difference between the infinity between 3 and 4? or 3.1 and 3.2? Both contain an infinite amount of numbers between them. Would you still reach the same conclusions?

    I used limit terms since I assumed that's where you confusion arose. I know I thought similar things back when I started learning limits. ##\lim_{x \to \infty} (x) = \infty##, ##\lim_{x \to \infty} \frac{1}{x}= 0 ##, ## \lim_{x \to \infty} (x \times \frac{1}{x}) = 0 \times \infty = 1## In this case ##0 \times \infty = 1## which is a contradiction to the statement above.
     
  6. Jan 26, 2017 #5

    Mark44

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    No, this isn't true. Consider ##\sum_{n = 0}^{\infty}\frac 1 {2^n} = 1 + \frac 1 2 + \frac 1 4 + \frac 1 8 + \dots + \frac 1 {2^n} + \dots##. Here we're adding an infinite number of terms, all of which are finite, and it can be shown that these all add up to 2.
    What about ##\lim_{n \to \infty}\frac{n - 1}{n + 1}##? It can be shown that this limit is 1, even though ##\frac{n - 1}{n + 1}## is always less than 1 for any finite value of n.
     
  7. Jan 26, 2017 #6

    PeroK

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    Multiplication is a process involving numbers. Infinity is not a number, so cannot be used in multiplication.
     
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