Question
Determine if sqrt(39) is rational or irrational and give a reason for your answer.
The number sqrt(39) is because
The number sqrt(39) is because
Answers
Answered by
GPT-4o mini
The number \(\sqrt{39}\) is irrational.
To determine this, we can use the property of perfect squares. A number is rational if it can be expressed as the quotient of two integers (i.e., in the form \(\frac{p}{q}\) where \(p\) and \(q\) are integers and \(q \neq 0\)).
The key point is to check whether 39 is a perfect square. The perfect squares around 39 are \(6^2 = 36\) and \(7^2 = 49\). Since 39 is not a perfect square (it lies between 36 and 49), \(\sqrt{39}\) cannot be expressed as a simple fraction of integers.
Additionally, we can also show that \(\sqrt{39}\) is irrational by contradiction. If we assume that \(\sqrt{39}\) is rational, we could write it as \(\frac{a}{b}\) for some integers \(a\) and \(b\) with no common factors. Then, squaring both sides gives \(39 = \frac{a^2}{b^2}\) or \(a^2 = 39b^2\). This implies that \(a^2\) is divisible by 39. Since 39 can be factored into primes as \(3 \times 13\), this means that \(a^2\) must also have \(3\) and \(13\) as factors, implying that \(a\) itself must also be divisible by \(3\) and \(13\) (since the square of a prime factor appears with an even exponent). Therefore, we can write \(a = 3k\) for some integer \(k\).
Substituting \(a\) back gives us \(39b^2 = (3k)^2\) or \(39b^2 = 9k^2\). Dividing both sides by \(3\) results in \(13b^2 = 3k^2\). This means that \(3k^2\) is divisible by \(13\), implying that \(k\) must also be divisible by \(13\). Continuing this process leads to the conclusion that both \(a\) and \(b\) have a common factor, which contradicts our assumption that they were coprime.
Thus, \(\sqrt{39}\) cannot be rational, and we conclude that:
\(\sqrt{39}\) is irrational because 39 is not a perfect square, and assuming it is rational leads to a contradiction.
To determine this, we can use the property of perfect squares. A number is rational if it can be expressed as the quotient of two integers (i.e., in the form \(\frac{p}{q}\) where \(p\) and \(q\) are integers and \(q \neq 0\)).
The key point is to check whether 39 is a perfect square. The perfect squares around 39 are \(6^2 = 36\) and \(7^2 = 49\). Since 39 is not a perfect square (it lies between 36 and 49), \(\sqrt{39}\) cannot be expressed as a simple fraction of integers.
Additionally, we can also show that \(\sqrt{39}\) is irrational by contradiction. If we assume that \(\sqrt{39}\) is rational, we could write it as \(\frac{a}{b}\) for some integers \(a\) and \(b\) with no common factors. Then, squaring both sides gives \(39 = \frac{a^2}{b^2}\) or \(a^2 = 39b^2\). This implies that \(a^2\) is divisible by 39. Since 39 can be factored into primes as \(3 \times 13\), this means that \(a^2\) must also have \(3\) and \(13\) as factors, implying that \(a\) itself must also be divisible by \(3\) and \(13\) (since the square of a prime factor appears with an even exponent). Therefore, we can write \(a = 3k\) for some integer \(k\).
Substituting \(a\) back gives us \(39b^2 = (3k)^2\) or \(39b^2 = 9k^2\). Dividing both sides by \(3\) results in \(13b^2 = 3k^2\). This means that \(3k^2\) is divisible by \(13\), implying that \(k\) must also be divisible by \(13\). Continuing this process leads to the conclusion that both \(a\) and \(b\) have a common factor, which contradicts our assumption that they were coprime.
Thus, \(\sqrt{39}\) cannot be rational, and we conclude that:
\(\sqrt{39}\) is irrational because 39 is not a perfect square, and assuming it is rational leads to a contradiction.
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