To prove that \( 34 + \sqrt{10} \) is irrational, we can begin by recalling a few definitions:
- A rational number is any number that can be expressed as the quotient of two integers, \( \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \).
- An irrational number is a number that cannot be expressed as a quotient of two integers.
Now we know that \( 34 \) is a rational number, since it can be expressed as \( \frac{34}{1} \).
Next, we analyze \( \sqrt{10} \):
- \( \sqrt{10} \) is known to be irrational. This can be shown by contradiction: if \( \sqrt{10} \) were rational, it could be expressed as \( \frac{p}{q} \) (in simplest form), leading to \( 10 = \frac{p^2}{q^2} \), or \( p^2 = 10q^2 \). This implies \( p^2 \) is divisible by 10, thus \( p \) must be at least divisible by \( \sqrt{10} \), which cannot happen with integers (since 10 is not a perfect square). Hence, \( \sqrt{10} \) cannot be rational.
Now, we combine these two components:
Suppose \( 34 + \sqrt{10} \) is rational. Then it can be expressed as \( \frac{m}{n} \) where \( m \) and \( n \) are integers. Rearranging gives us: \[ \sqrt{10} = \frac{m}{n} - 34 \] This expression indicates that \( \sqrt{10} \) is equal to a rational number (the right-hand side). However, since \( \sqrt{10} \) is irrational, this leads to a contradiction. Thus, our assumption that \( 34 + \sqrt{10} \) is rational must be false.
Therefore, \( 34 + \sqrt{10} \) is irrational.
Conclusion
In summary, the sum \( 34 + \sqrt{10} \) is irrational because it combines the rational number 34 with the irrational number \( \sqrt{10} \), and the sum of a rational and an irrational number is always irrational.
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