Prove that 3.999... = 4
(In reply to
Then... by Dulanjana)
Yes, but not as a different number from 4. Between any two numbers it is possible to find "at least one" (actually an infinite number) of intermediate numbers. But with 3.999..., if you were to add o.ooo... 00100... (with the 1 in any space but the "last" space), the resulting number would be greater than 4, not between 3.9999.... and 4
On the other hand, 3.999... is useful as an alternate way of writing 4, and similar notation for other whole numbers makes sense of the following sum: 1/3 + 2/3 = .3333... + .6666... = .9999...
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Posted by TomM
on 2002-06-08 12:36:07 |