Question
Prove that the sum of 3/4
and radical notation 10
is irrational. Show your work, and in 1–2 sentences, explain why the sum is an irrational number.
and radical notation 10
is irrational. Show your work, and in 1–2 sentences, explain why the sum is an irrational number.
Answers
Answered by
GPT-4o mini
To prove that the sum of \( \frac{3}{4} \) and \( \sqrt{10} \) is irrational, we will first establish that \( \sqrt{10} \) is an irrational number.
### Step 1: Show that \( \sqrt{10} \) is irrational
Assume, for the sake of contradiction, that \( \sqrt{10} \) is rational. This means it can be expressed as \( \frac{a}{b} \), where \( a \) and \( b \) are integers, \( b \neq 0 \), and \( \gcd(a, b) = 1 \). Therefore, we have:
\[
\sqrt{10} = \frac{a}{b}
\]
Squaring both sides gives:
\[
10 = \frac{a^2}{b^2} \implies a^2 = 10b^2
\]
This implies that \( a^2 \) is divisible by 10. Since 10 can be factored into \( 2 \times 5 \), it follows that \( a^2 \) must also be divisible by both 2 and 5, meaning \( a \) must be divisible by both 2 and 5 (since a square of an integer is divisible by a prime if and only if the integer itself is divisible by that prime). Therefore, let \( a = 10k \) for some integer \( k \).
Substituting back, we get:
\[
a^2 = (10k)^2 = 100k^2 \implies 100k^2 = 10b^2 \implies 10k^2 = b^2
\]
This shows \( b^2 \) is also divisible by 10, so \( b \) must also be divisible by both 2 and 5. This contradicts our assumption that \( a \) and \( b \) are coprime. Thus, \( \sqrt{10} \) is irrational.
### Step 2: Show that \( \frac{3}{4} + \sqrt{10} \) is irrational
Now we know \( \sqrt{10} \) is irrational. Next, we will show that the sum of a rational number \( \frac{3}{4} \) and an irrational number \( \sqrt{10} \) is irrational.
Assume, for contradiction, that \( \frac{3}{4} + \sqrt{10} \) is rational. This means it can be expressed as:
\[
\frac{3}{4} + \sqrt{10} = \frac{m}{n}
\]
for some integers \( m \) and \( n \) with \( n \neq 0 \). We can isolate \( \sqrt{10} \):
\[
\sqrt{10} = \frac{m}{n} - \frac{3}{4} = \frac{4m - 3n}{4n}
\]
Since \( \frac{4m - 3n}{4n} \) is a rational number (as both numerator and denominator are integers), it follows that \( \sqrt{10} \) is rational, which contradicts our earlier conclusion.
### Conclusion
Thus, \( \frac{3}{4} + \sqrt{10} \) must be irrational.
**Explanation**: The sum of a rational number (\( \frac{3}{4} \)) and an irrational number (\( \sqrt{10} \)) is always irrational because it is impossible for the sum to result in a rational number without contradicting the nature of the irrational component.
### Step 1: Show that \( \sqrt{10} \) is irrational
Assume, for the sake of contradiction, that \( \sqrt{10} \) is rational. This means it can be expressed as \( \frac{a}{b} \), where \( a \) and \( b \) are integers, \( b \neq 0 \), and \( \gcd(a, b) = 1 \). Therefore, we have:
\[
\sqrt{10} = \frac{a}{b}
\]
Squaring both sides gives:
\[
10 = \frac{a^2}{b^2} \implies a^2 = 10b^2
\]
This implies that \( a^2 \) is divisible by 10. Since 10 can be factored into \( 2 \times 5 \), it follows that \( a^2 \) must also be divisible by both 2 and 5, meaning \( a \) must be divisible by both 2 and 5 (since a square of an integer is divisible by a prime if and only if the integer itself is divisible by that prime). Therefore, let \( a = 10k \) for some integer \( k \).
Substituting back, we get:
\[
a^2 = (10k)^2 = 100k^2 \implies 100k^2 = 10b^2 \implies 10k^2 = b^2
\]
This shows \( b^2 \) is also divisible by 10, so \( b \) must also be divisible by both 2 and 5. This contradicts our assumption that \( a \) and \( b \) are coprime. Thus, \( \sqrt{10} \) is irrational.
### Step 2: Show that \( \frac{3}{4} + \sqrt{10} \) is irrational
Now we know \( \sqrt{10} \) is irrational. Next, we will show that the sum of a rational number \( \frac{3}{4} \) and an irrational number \( \sqrt{10} \) is irrational.
Assume, for contradiction, that \( \frac{3}{4} + \sqrt{10} \) is rational. This means it can be expressed as:
\[
\frac{3}{4} + \sqrt{10} = \frac{m}{n}
\]
for some integers \( m \) and \( n \) with \( n \neq 0 \). We can isolate \( \sqrt{10} \):
\[
\sqrt{10} = \frac{m}{n} - \frac{3}{4} = \frac{4m - 3n}{4n}
\]
Since \( \frac{4m - 3n}{4n} \) is a rational number (as both numerator and denominator are integers), it follows that \( \sqrt{10} \) is rational, which contradicts our earlier conclusion.
### Conclusion
Thus, \( \frac{3}{4} + \sqrt{10} \) must be irrational.
**Explanation**: The sum of a rational number (\( \frac{3}{4} \)) and an irrational number (\( \sqrt{10} \)) is always irrational because it is impossible for the sum to result in a rational number without contradicting the nature of the irrational component.