Does 0.99999... really equal 1?
Does 0.999 (repeated) equal "one"? You might be surprised by the answer... which is "yes". I heard this on a couple of podcasts over the last few months, and I still find it hard to accept at face value. However, I am now convinced and here's several simple proofs that .99999 does indeed equal 1...
Method 1:
Let x = 0.9999...
Then 10x = 9.9999...
If we then subtract x from both sides of the equation, then:
10x - x = 9.9999... - 0.9999...
So, 9x = 9
Divide both sides of the equation by 9, and...
x = 1 ... which, when we started, we said = 0.9999...
Method 2:
1/9 = 0.11111...
Multiply both sides of the equation by 9:
9 X 1/9 = 9 X 0.11111...
1 = 0.99999...
Method 3:
We know that 0.9 is not equal to 1; neither is 0.999, nor 0.99999. If you stop the expansion of 9s at any finite point, the fraction you have (like .99999 = 99999/100000) is never equal to 1. But each time you add a 9, the margin error is smaller (with each 9, the error is actually ten times smaller).
You can show (using calculus or other summations) that with a large enough number of 9s in the expansion, you can get arbitrarily close to 1. There is no other number that the sequence gets arbitrarily close to - it is always 1. Another way of saying this is that "the limit is 1".
Thus, if you are going to assign a value to 0.9999..., the only sensible value is "1".
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I love the empirical truth you have uncovered and exposed. I deal in.0000's everyday at work so I understand the need for truth in numbers. And I believe that numbers are the only universal language and slang makes people look stupid.
ok, i know a couple other ways.- firstly, 1/3 is equal to 0.333333333333333.... correct? so, if 1/3=0.33333...., multiply both sides by 3. you get 1=0.99999999..... Another way is that. if you put a repeating decimal's number like this over 9.. for example 1/3= 0.333.. so put the 3 over 9. therefore... 0.3333...=1/3. So, if you put 0.999.... you put 9 over 9 you get 9/9 which is equal to 1.