.999... = 1

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[Serious]
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#1

Post by [Serious] »

It just does.
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seeker
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#2

Post by seeker »

Practically speaking I would agree but if we are being strictly mathematical we are talking about an infinite sequence bounded by one but not including one so, strictly speaking, it never truly equals 1. One way to think of it is for whatever number in the set (.9999..n) there is going to be a number (.9999...n+1) between it and the absolute value of 1. Pedantic? Sure but depending on context it can be important.
[Serious]
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#3

Post by [Serious] »

seeker wrote: Mon Aug 08, 2022 7:32 pm Practically speaking I would agree but if we are being strictly mathematical we are talking about an infinite sequence bounded by one but not including one so, strictly speaking, it never truly equals 1. One way to think of it is for whatever number in the set (.9999..n) there is going to be a number (.9999...n+1) between it and the absolute value of 1. Pedantic? Sure but depending on context it can be important.
Actually, from a strictly mathematical standpoint it's exactly equal to 1.

Starting out with the common proof:
A=.9999....
10a= 9.9999...
9a= 9
A=1
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