prove that between two rational number there is an irrational
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.
To prove that between two rational numbers there is an irrational number, we can use the concept of density of the real numbers.
Let's assume that we have two rational numbers, a and b, where a < b. We want to find an irrational number x such that a < x < b.
To begin, let's consider the average of a and b, denoted as (a + b)/2.
Case 1: (a + b)/2 is irrational
If (a + b)/2 is already irrational, then we have found the irrational number x such that a < x < b.
Case 2: (a + b)/2 is rational
If (a + b)/2 is rational, we can use the concept of density to find an irrational number between a and (a + b)/2, and another one between (a + b)/2 and b.
1. Between a and (a + b)/2:
- We know that (a + b)/2 is rational.
- The decimal representation of a rational number can either terminate or repeat.
- Therefore, there are infinitely many irrational numbers between a and (a + b)/2. We can choose one of them as x1.
2. Between (a + b)/2 and b:
- We can repeat the same logic as above.
- There are infinitely many irrational numbers between (a + b)/2 and b. We can choose one of them as x2.
Therefore, we have found two irrational numbers, x1 and x2, such that a < x1 < (a + b)/2 < x2 < b.
In conclusion, between any two rational numbers, there exists at least one irrational number.