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# 0.999999999999999999999999999999999999999999999999999999999999999999.... equal 1? (Art of Mathematics question on Youtube)

## 0.999999999999999999999999999999999999999999999999999999999999999999.... equal 1? (Art of Mathematics question on Youtube)

Seva Lamberdar

Posts : 6575
Join date : 2012-11-29

## Does 0.999999999.... equal 1? (Art of Mathematics question on Youtube)

It reminds of the analogy about parallel lines meeting at infinity only, implying that infinity (∞) is only an imaginary idea (with ∞ as indeterminate).

Thus, mathematically, 1/∞ = 0 is an assumption (approximation), while replacing the inexact (imprecise) 1/∞  with exact (precise) 0.

Let's work out a solution in the following  with the implicit use of   1/∞ as 0  to show 0.99999999999.. as approaching 1 :

We can write 0.9999999999 .... (to nth term in 9) as

S = 0.9 + 0.09 + 0.009 + 0.0009 + 0.00009 + ... + (nth term), thereby

S = 9 (1/10 + 1/100 + 1/1000 + 1/10000 +.... + nth term), or

S = 9*A ....... Eqn. (i),

where A  = 1/10 + 1/100 + 1/1000 +  ............. + 1/(nth power in 10), or

A = (10)^(-1) + (10)^(-2) + (10)^(-3) + ......  + (10)^(-n),  ........ Eqn. (ii)

Dividing both sides of Eqn. (ii) with 10 yields the following,

A/10 = (10)^(-2) + (10)^(-3) + ............. + (10)^(-n) + (10)^(-n-1)  --- Eqn. (iii)

Subtracting Eqn (iii) from Eqn. (ii) results in the following,

A - A/10 = (10)^(-1) - (10)^(-n-1), leading to

9A/10 = (1/10)*(1 - 1/(10^n)), or

A = (1/9 )*(1 - 1/(10^n))  ....   Eqn. (iv).

Substituting A from Eqn. (iv) in Eqn (i) yields,

S = 9*(1/9)*(1 - 1/(10^n)), or

S = 1 - 1/10^n ........  Eqn. (v).

In Eqn. (v), S (or 0.99999999999999..) →1, as 1/10^n → 0 for large n, while using the assumption that (inexact) 1/∞  → (exact) 0.

Seva Lamberdar

Posts : 6575
Join date : 2012-11-29

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