Question
Prove that the sum of 3/4 in radical 10 is irrational show your work
Answers
GPT-4o mini
To show that the sum \( \frac{3}{4} + \sqrt{10} \) is irrational, we will use a proof by contradiction.
**Assumption:** Suppose \( \frac{3}{4} + \sqrt{10} \) is rational. Then we can express it as:
\[
\frac{3}{4} + \sqrt{10} = r
\]
for some rational number \( r \).
**Rearranging the equation:**
\[
\sqrt{10} = r - \frac{3}{4}
\]
Next, let's express \( r \) in a way that makes this clearer. Since \( r \) is rational, it can be written as a fraction \( \frac{a}{b} \), where \( a \) and \( b \) are integers (and \( b \neq 0 \)). Thus:
\[
\sqrt{10} = \frac{a}{b} - \frac{3}{4}
\]
To combine these fractions, we need a common denominator, which is \( 4b \):
\[
\sqrt{10} = \frac{4a - 3b}{4b}
\]
**Now, observe the properties of each side:**
1. **Rationality of \( \frac{4a - 3b}{4b} \):** Since \( a \) and \( b \) are integers, \( 4a - 3b \) and \( 4b \) are also integers. Therefore, the right side \( \frac{4a - 3b}{4b} \) is rational.
2. **Irrationality of \( \sqrt{10} \):** We know that \( \sqrt{10} \) is irrational.
**Contradiction:** We have reached a contradiction because we have expressed \( \sqrt{10} \) as a rational number (on the right side) while simultaneously knowing it is irrational.
Thus, our original assumption that \( \frac{3}{4} + \sqrt{10} \) is rational must be false.
Therefore, we conclude that
\[
\frac{3}{4} + \sqrt{10} \text{ is irrational.}
\]
This completes the proof.
**Assumption:** Suppose \( \frac{3}{4} + \sqrt{10} \) is rational. Then we can express it as:
\[
\frac{3}{4} + \sqrt{10} = r
\]
for some rational number \( r \).
**Rearranging the equation:**
\[
\sqrt{10} = r - \frac{3}{4}
\]
Next, let's express \( r \) in a way that makes this clearer. Since \( r \) is rational, it can be written as a fraction \( \frac{a}{b} \), where \( a \) and \( b \) are integers (and \( b \neq 0 \)). Thus:
\[
\sqrt{10} = \frac{a}{b} - \frac{3}{4}
\]
To combine these fractions, we need a common denominator, which is \( 4b \):
\[
\sqrt{10} = \frac{4a - 3b}{4b}
\]
**Now, observe the properties of each side:**
1. **Rationality of \( \frac{4a - 3b}{4b} \):** Since \( a \) and \( b \) are integers, \( 4a - 3b \) and \( 4b \) are also integers. Therefore, the right side \( \frac{4a - 3b}{4b} \) is rational.
2. **Irrationality of \( \sqrt{10} \):** We know that \( \sqrt{10} \) is irrational.
**Contradiction:** We have reached a contradiction because we have expressed \( \sqrt{10} \) as a rational number (on the right side) while simultaneously knowing it is irrational.
Thus, our original assumption that \( \frac{3}{4} + \sqrt{10} \) is rational must be false.
Therefore, we conclude that
\[
\frac{3}{4} + \sqrt{10} \text{ is irrational.}
\]
This completes the proof.