Prove or disprove 0.0000.....001 equals 0

[BMAD]

Prove or disprove that 0.000....000001 equals 0

[bonanova]

  Reveal hidden contents

This question appears similar to the one that asks for a proof that 0.9999... = 1 which is paradoxical until we establish that the notation "..." stands for infinite repetition. That is, the equality holds only if there is no "last" nine: they go on forever. But then the expression given in the OP has no clear meaning. It implicitly claims that after an infinite number of zeros there can be (a) five more zeros and then (b) a final unity. There can be neither. The infinity represented by "..." has the lowest cardinality of all the infinities, but it is infinity nonetheless. Nothing follows it.

If we change the meaning of "..." we might say that 0.000...000001 reads: limit (n->inf) [0.000 followed by n zeros followed by five zeros followed by 1]. In that case, the expression evaluates to zero and the proof is simple. The expression is positive, decreasing and bounded below by zero. For any positive epsilon, no matter how small, there is an n for which the expression is (positive and) smaller than epsilon.

 

[BMAD]

  On 1/21/2017 at 4:05 PM, bonanova said:

  Reveal hidden contents

This question appears similar to the one that asks for a proof that 0.9999... = 1 which is paradoxical until we establish that the notation "..." stands for infinite repetition. That is, the equality holds only if there is no "last" nine: they go on forever. But then the expression given in the OP has no clear meaning. It implicitly claims that after an infinite number of zeros there can be (a) five more zeros and then (b) a final unity. There can be neither. The infinity represented by "..." has the lowest cardinality of all the infinities, but it is infinity nonetheless. Nothing follows it.

If we change the meaning of "..." we might say that 0.000...000001 reads: limit (n->inf) [0.000 followed by n zeros followed by five zeros followed by 1]. In that case, the expression evaluates to zero and the proof is simple. The expression is positive, decreasing and bounded below by zero. For any positive epsilon, no matter how small, there is an n for which the expression is (positive and) smaller than epsilon.

 

Expand  

One of my students used this as their proof, I am curious about what you think of it bonanova,

Assume that 0.000....1 exists then 1 - 0.000....1 = 0.999....

Since we can show 0.99999 = 1  and 0.000...1 + 0.999... = 1 then 0.000...1 must equal zero as 0+1=1.

[bonanova]

Well the issue is what 0,000...1 means. That question must be answered before asking whether it has a value, and if so what the value is.

The expression contains a string that is both infinite and terminating. A terminating string can be counted. Each digit occupies a numbered place. What place does the 1 occupy? This expression implies that a finite string that terminates with a 1 can contain an infinite sub-string of zeros. That fact is equivalent to saying that any explanation of the meaning of the expression must contain a contradiction.

We can construct the sequence

• 0.1
• 0.01
• 0.001
• 0.0001
• 0.00001
• ...

and ask whether the sequence converges. Yes. It easily can be proved to converge to zero. But the proof never mentions 0.000...1. So it does not apply to the OP.

Regarding your student's proof I question the assertion 1 - 0.000....1 = 0.999.... Why? Because it equates an infinite string to a terminating string.

I would say 1 - 0.000....1   =   0.999... - 0.000...1   =   0.999...8 whose meaning is equally problematical. Saying that it evaluates to unity carries the same difficulty as the original question.