Multiplication problem

Question

I have a very simple question

As 0.999⋯≠1

Hence,

var one = 1;
var less_than_one = 0.99999999999999999 (let's say to a computable decimal point)
var greater_than_one = 1.00000000000000000000000.....1 (let's say to a computable decimal point)
var n1 = 10;
var n2 = 10;
var n3 = 10;
for i = 0 to to infinity (let's say a large computable number)
    n1 = n1 * one;
    n2 = n2 * less_than_one;
    n3 = n3 * greater_than_one;

print (n1);
print (n2);
print (n3);

Result :

n1 = 10
n2 = 0.000000000000000 (tends to zero, as will considered by many computers)
n3 = A larger number

I understand the basic principle behind why this is happening, as any number multiplied by less than one will result in a value less than the number multiplied.

Now, why I am asking this question here is because I have been asked by a 6th-grade nephew when he runs this with a computer language and now I have to explain why this happening to him back.

I tried but he is not satisfied,

Could you please help me explain to him why this is happening with real life problems?

Edit1: Please fell free to edit and add a proper tag as I am new to the community.

Answer 1

No.

$0.999999999999 = 1$

Assume it.

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Clearly: $$ 0.999999=1-0.000001 $$

Well, this is because the remainder is tending to zero, so as long you add 9's you make the series aproach more and more to 1. In the ideal limit, which we write this number as $0.\overline9$, they are the same: $$ 0.9999999...=0.\overline9 = \lim_{1 \to \infty}1 - \frac{1}{n} =1 $$

In real life, you can explain the Zenon paradox.. what happens if you walk the half of the path, and then the remaining half, and then the remaining half, and then the remaining.....

Answer 2

You essentially want to show that if $0<r<1$ then $r^k \to 0$ as $k \to \infty$.

As you mentioned already, $r,r^2,r^3,\ldots$ is a strictly decreasing sequence. If you can show that eventually this sequence becomes less than $\epsilon$ for any $0<\epsilon < 1$, then the limit must be zero. To see this, note that for $K = \lceil\log(\epsilon)/\log(r)\rceil$ we have $r^K \le \epsilon$.