Why .99999=1

This game is the definition of a limit. As it turns out, all limits are proof by contradiction. We see that as x goes to infinity lt1 f x remains 1, while lt1 1 is 0, which suggests that, even in the limit of infinite nines,.

Taking a step back, your argument uses limits about a point, and indeed a function can have a discontinuity about that point. I am talking about infinite limits, which have a slightly different definition.

But instead of getting into the details of the difference in the definition of a limit about a point and an infinite limit, let me point blank ask you: What does 0. What is it short hand for? Mathematically that means 0. So I the infinite definition of a limit is the appropriate way to understand it.

I am passionate about this since this is one of the facts that could really help people understand infinite limits. How Can 0. There are many different proofs of the fact that " 0. So why does this question keep coming up? Students don't generally argue with " 0. Maybe it's just that it "feels" "wrong" that something as nice and neat and well-behaved as the number " 1 " could also be written in such a messy form as " 0.

One of the major sticking points seems to be notational, so let me get that out of the way first: When I say " 0. The ellipsis the "dot, dot, dot" after the last 9 means "goes on forever in like manner".

In other words, " 0. There will always be another " 9 " to tack onto the end of 0. So don't object to 0. Yes, at any given stop, at any given stage of the expansion, for any given finite number of 9 s, there will be a difference between 0. But the point of the "dot, dot, dot" is that there is no end; 0. There is no "last" digit. So the "there's always a difference" argument betrays a lack of understanding of the infinite. That's not a "criticism", per se; infinity is a messy topic.

Proof by geometric series. The number " 0. In other words, each term in this endless summation will have a " 9 " preceded by some number of zeroes. This may also be written as:. Since the size of the common ratio r is less than 1 , we can use the infinite-sum formula to find the value:. So the formula proves that 0. Note: Technically, the above proof requires that some fairly advanced concepts be taken on faith.

If you study "foundations" or mathematical philosophy way after calculus , you may encounter the requisite theoretical constructs. Other pre-calculus arguments. Reasonably then, 0. But 3 0. Then 0. Argument from arithmetic: When you subtract a number from itself, the result is zero. So what is the result when you subtract 0.