Does 0.99999999999999999999999999999999999999999 repeating equals 1?
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51-55, F
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@Invisible infinitesimal math deals with the number between .999... and 1. The "problem" with your math is that 9/10 is a defined number. .999... is an ill defined number because it goes off into infinity, which doesn't exist.
you can add defined numbers (9/10) + (9/100) + (9/1000) and so on forever. It will never get you to 1.
For most purposes does .999... equal 1? Sure. Is there a branch of math that says .000...1 exists? Yes.
you can add defined numbers (9/10) + (9/100) + (9/1000) and so on forever. It will never get you to 1.
For most purposes does .999... equal 1? Sure. Is there a branch of math that says .000...1 exists? Yes.
JoyfulSilence · 46-50, M
Yes.
The value 0.999999999999999999...
is the limit, as n goes to infinity, of a sequence of values S(n), where:
S(n) = Sum from k=1..n of 9 * (0.1)^k
This sequence S(n) converges to the limit 1.
Proof:
We know that
S(n) = Sum from k=1..n of 9 * (0.1)^k
= 9 * Sum from k=1..n of (0.1)^k
= 9 * { [ Sum from k=0..n of (0.1)^k ] - 1 }
It can be shown that
Sum from k=0..n of (0.1)^k
= [ 1 - (0.1)^(n+1) ] / [ 1 - 0.1 ]
= (10/9) * [ 1 - (0.1)^(n+1) ]
So
S(n) = 9 * { (10/9) * [ 1 - (0.1)^(n+1) ] - 1 }
= 10 * [ 1 - (0.1)^(n+1) ] - 9
As n increases the term (0.1)^(n+1) goes to zero.
Hence, S(n) converges to
= 10 * [ 1 - 0 ] - 9
= 1
To prove that
Sum from k=0..n of (0.1)^k
= [ 1 - (0.1)^(n+1) ] / [ 1 - 0.1 ]
First, let a = 0.1. Use induction.
If n = 0, then
Sum from k=0..n of a^k
= Sum from k=0..0 of a^k
= a^0
= 1
= [ 1 - a ] / [ 1 - a ]
= [ 1 - a^(0+1) ] / [ 1 - a ]
= [ 1 - a^(n+1) ] / [ 1 - a ]
Now, assume that
Sum from k=0..n of a^k
= [ 1 - a^(n+1) ] / [ 1 - a ]
and show that
Sum from k=0..(n+1) of a^k
= [ 1 - a^(n+2) ] / [ 1 - a ]
Well, we have:
Sum from k=0..(n+1) of a^k
= { Sum from k=0..n of a^k } + a^(n+1)
= [ 1 - a^(n+1) ] / [ 1 - a ] + a^(n+1)
= { 1 - a^(n+1) + [ 1 - a ] * a^(n+1) } / [ 1 - a ]
= [ 1 - a * a^(n+1) ] / [ 1 - a ]
= [ 1 - a^(n+2) ] / [ 1 - a ]
QED
The value 0.999999999999999999...
is the limit, as n goes to infinity, of a sequence of values S(n), where:
S(n) = Sum from k=1..n of 9 * (0.1)^k
This sequence S(n) converges to the limit 1.
Proof:
We know that
S(n) = Sum from k=1..n of 9 * (0.1)^k
= 9 * Sum from k=1..n of (0.1)^k
= 9 * { [ Sum from k=0..n of (0.1)^k ] - 1 }
It can be shown that
Sum from k=0..n of (0.1)^k
= [ 1 - (0.1)^(n+1) ] / [ 1 - 0.1 ]
= (10/9) * [ 1 - (0.1)^(n+1) ]
So
S(n) = 9 * { (10/9) * [ 1 - (0.1)^(n+1) ] - 1 }
= 10 * [ 1 - (0.1)^(n+1) ] - 9
As n increases the term (0.1)^(n+1) goes to zero.
Hence, S(n) converges to
= 10 * [ 1 - 0 ] - 9
= 1
To prove that
Sum from k=0..n of (0.1)^k
= [ 1 - (0.1)^(n+1) ] / [ 1 - 0.1 ]
First, let a = 0.1. Use induction.
If n = 0, then
Sum from k=0..n of a^k
= Sum from k=0..0 of a^k
= a^0
= 1
= [ 1 - a ] / [ 1 - a ]
= [ 1 - a^(0+1) ] / [ 1 - a ]
= [ 1 - a^(n+1) ] / [ 1 - a ]
Now, assume that
Sum from k=0..n of a^k
= [ 1 - a^(n+1) ] / [ 1 - a ]
and show that
Sum from k=0..(n+1) of a^k
= [ 1 - a^(n+2) ] / [ 1 - a ]
Well, we have:
Sum from k=0..(n+1) of a^k
= { Sum from k=0..n of a^k } + a^(n+1)
= [ 1 - a^(n+1) ] / [ 1 - a ] + a^(n+1)
= { 1 - a^(n+1) + [ 1 - a ] * a^(n+1) } / [ 1 - a ]
= [ 1 - a * a^(n+1) ] / [ 1 - a ]
= [ 1 - a^(n+2) ] / [ 1 - a ]
QED
Deadcutie · 18-21, F
No.. .9 will never be 1