The square root of 73, denoted as \( \sqrt{73} \), is considered an irrational number. This can be reasoned as follows:
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Definition of Rational Numbers: A number is rational if it can be expressed as a fraction of two integers (i.e., in the form \( \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \)).
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Perfect Squares: A rational square root occurs only for perfect squares (numbers that can be expressed as the square of an integer).
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Checking for Perfect Squares: The perfect squares closest to 73 are \( 64 \) (which is \( 8^2 \)) and \( 81 \) (which is \( 9^2 \)). Since 73 is not equal to the square of any integer, it is not a perfect square.
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Irrationality Proof: More formally, one can prove that \( \sqrt{73} \) is irrational by contradiction:
- Assume \( \sqrt{73} \) is rational, meaning we can express it as \( \frac{p}{q} \) where \( p \) and \( q \) are integers with no common factors (i.e., in lowest terms).
- Then, \( 73 = \frac{p^2}{q^2} \) which implies \( p^2 = 73q^2 \).
- This means \( p^2 \) is divisible by 73, and hence \( p \) must also be divisible by 73 (since 73 is prime).
- Let \( p = 73k \) for some integer \( k \), substituting this back gives \( (73k)^2 = 73q^2 \) or \( 5329k^2 = 73q^2 \).
- Dividing through by 73 results in \( 73k^2 = q^2 \), showing \( q^2 \) is also divisible by 73, thus \( q \) must be divisible by 73.
- This means both \( p \) and \( q \) share a factor of 73, contradicting our initial assumption that they had no common factors.
Since we reach a contradiction, we conclude that \( \sqrt{73} \) cannot be expressed as a fraction of two integers, and therefore, it is irrational.
Thus, \( \sqrt{73} \) is irrational.