prove that between two rational number there is an irrational

Answer 1

proof:

Suppose for contradiction that between any two rational numbers there is no irrational number. Let a and b be any two distinct rational numbers, where without loss of generality we can assume that a < b.

Then, according to our assumption, there can be no irrational number x such that a < x < b. However, we know that the difference b - a is rational (since a and b are both rational), and that this difference is positive: b - a > 0.

Now consider the number (b - a) / √2. Since √2 is irrational, we already know that this number is not rational. But we can also see that:

a + ((b - a) / √2) < b

To see this, note that (b - a) / √2 is positive (since b > a) and that adding a positive number to a makes it bigger.

So we have shown that there exists an irrational number x such that a < x < b, which contradicts our assumption. Therefore, our assumption must be false, and we conclude that between any two rational numbers there must be at least one irrational number.