To prove that the sum \( \frac{3}{4} + \sqrt{10} \) is irrational, we can use proof by contradiction.
-
Assume the contrary: Suppose \( \frac{3}{4} + \sqrt{10} \) is rational. This means that we can express it as a fraction of two integers, say \( r \) where \( r = \frac{p}{q} \) for some integers \( p \) and \( q \) with \( q \neq 0 \).
-
Isolate \( \sqrt{10} \): Rearranging gives: \[ \sqrt{10} = r - \frac{3}{4} \] Rewriting \( r \) in terms of a common denominator: \[ \sqrt{10} = \frac{p}{q} - \frac{3}{4} = \frac{4p - 3q}{4q} \]
-
Since the right side \( \frac{4p - 3q}{4q} \) is a fraction of integers, it follows that \( \sqrt{10} \) would also have to be rational because the left side is a rational expression.
-
Rationality of \( \sqrt{10} \): However, \( \sqrt{10} \) is known to be irrational. We can prove this by noting that 10 is not a perfect square, and thus \( \sqrt{10} \) cannot be expressed as a fraction \( \frac{a}{b} \) where \( a \) and \( b \) are integers with no common factors (that is, \( \sqrt{10} \) cannot be simplified to a rational number).
-
Conclusion: Since our assumption that \( \frac{3}{4} + \sqrt{10} \) is rational leads to the contradiction that \( \sqrt{10} \) must be rational, our initial assumption must be false. Therefore, \( \frac{3}{4} + \sqrt{10} \) is irrational.
In summary, the sum \( \frac{3}{4} + \sqrt{10} \) is irrational because it includes \( \sqrt{10} \), which is an irrational number, and the sum of a rational number and an irrational number is always irrational.
1 answer