Question

give the reason for the natural number to be discontinuous​

Answer 1

Answer:

There is a beautiful function, known as the Dirichlet function, that satisfies your condition.

f(x)∈Q ? 1:0

If x is rational, then f(x) = 1

If x is irrational, then f(x) = 0

But why is this discontinuous at all points? Let me explain.

Between two irrational numbers there is always a rational number.

Take any two irrational numbers, that you think are close together.

Being irrational, their decimal sequences never terminate.

Now take the greater number and traverse its digits from left to right, comparing to the corresponding digits of the lesser number. As soon as you encounter a digit that does not match the corresponding digit of the lesser number, chop off all digits after that one. Voila, you have obtained a rational number between the two irrational numbers.

For example, take the numbers:

2.124182731082741924612981322...

2.124182731082741924612996531...

Following the above algorithm, you get the number:

2.12418273108274192461299, which is rational.

Similarly, there is always an irrational number between two rational numbers.

Take any two rational numbers. Take the lesser one and give it a non-terminating non-repeating decimal sequence. You then, have an irrational number between the two rational numbers.

Coming back to the function at hand.

Let f be continuous for a point x and x + δ (where δ→0).

That means x and x + δ are either both irrational, or both rational.

But that means, there lies a value between x and x + δ, with opposite nature to them.

By contradiction, this makes the function discontinuous.

A simple, but elegant proof. :)

Step-by-step explanation:

hope it helps

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