prove that the sum of 3/4 and sqrt 10 is irrational

Answer 1

To prove that the sum \( \frac{3}{4} + \sqrt{10} \) is irrational, we can use a proof by contradiction.

1. Assume the contrary: Suppose \( \frac{3}{4} + \sqrt{10} \) is rational. This means we can express it as a fraction of two integers:

   \[ \frac{3}{4} + \sqrt{10} = r \]

   where \( r \) is a rational number.

2. Isolate \( \sqrt{10} \):

   Subtract \( \frac{3}{4} \) from both sides:

   \[ \sqrt{10} = r - \frac{3}{4} \]

3. Express \( r \) in a suitable form:

   Since \( r \) is rational, we can write it as \( r = \frac{a}{b} \) where \( a \) and \( b \) are integers and \( b \neq 0 \). Therefore,

   \[ \sqrt{10} = \frac{a}{b} - \frac{3}{4} \]

   To combine these fractions, we find a common denominator, which is \( 4b \):

   \[ \sqrt{10} = \frac{4a - 3b}{4b} \]

4. Rationality of \( \sqrt{10} \):

   Since \( 4a - 3b \) and \( 4b \) are both integers, the right side of the equation is a ratio of two integers. This means \( \sqrt{10} \) is rational.

5. Contradiction:

   However, we know that \( \sqrt{10} \) is irrational (as \( 10 \) is not a perfect square). Therefore, our initial assumption that \( \frac{3}{4} + \sqrt{10} \) is rational must be incorrect.

6. Conclusion:

   Hence, we conclude that \( \frac{3}{4} + \sqrt{10} \) is irrational.