Show that if x is any real number, there is a sequence of rational numbers converging to x.

Mathematics

1 answer:

Gnesinka [82]3 years ago

3 0

Answer:

It has been proven from the explanation.

Step-by-step explanation:

If we assume that the real number x is greater than 0. Now, If on the contrary, a is less than 0, we can make the argument that;

Let us make n to be a natural number and m = m(n) to be the largest positive integer in such a way that;

m/n < x

Thus, (m+1)/n ≥ x and finally;

|x - m/n| < 1/n

If r_n = m/n. It will be easy to show from the definition of limits that the sequence (r_n) has limit x.

From earlier where we assumed that x > 0, the numbers that will be obtained by truncating the decimal expansion of "x" at the n-th place will be rational, and clearly have the limit "x". Although it implies we are now assuming every real number will have a decimal expansion.

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Step-by-step explanation:

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