The spiraling construction method illustrated below uses the Pythagorean theorem to create the square roots of all of the integers from \(1\) onward. But despite the consistency of the construction, the numbers formed fall into two disjoint classes: the rational numbers and the irrational numbers.

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Rational numbers are those which can be represented as a fraction of two integers, having the form \(\frac{m}{n}\), where \(m\) and \(n\) are integers and \(n \ne 0\). Numbers such \(\frac{1}{2}\), \(\frac{3}{5}\), and \(\frac{14}{9}\) are rational numbers. The integers are also rational numbers, as they can be represented as themselves divided by \(1\) (eg. \(7 = \frac{7}{1}\)). The irrational numbers are numbers which cannot be represented this way. In their decimal form, they are represented by an infinite string of digits which do not exhibit any pattern. The most famous irrational number is perhaps \(\pi = 3.1415926…\), but others include Euler's number, \(e\), and the golden ratio, \(\phi\). Some of the square roots of numbers are irrational, including \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{20}\), \(\sqrt{130}\), and \(\sqrt{\frac{4}{5}}\), while other square roots of numbers are rational, such as \(\sqrt{4}\) = \(2\), \(\sqrt{9}\) = \(3\) \(\sqrt{144}\) = \(12\), \(\sqrt{\frac{9}{100}} = \frac{3}{10}\).

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A method for evaluating whether or not the square root of a number is irrational:

Reduce the number (or the numerator and the denominator of the number if the number is a fraction) under the square root into its prime factorization. If all of the prime factors have even exponents, then the square root of that number is rational. If one or more of the prime factors have odd exponents, then the square root of that number is irrational. Exploring \(\sqrt{144}\), the prime factorization of \(144\) is \(2^{4}3^{2}\). All of the prime factors, which in this case are \(2\) and \(3\), have even exponents, so \(\sqrt{144}\) is rational. Exploring \(\sqrt{20}\), the prime factorization of \(20\) is \(2^{2}5^{1}\). The prime factor \(5\) has an odd exponent, so \(\sqrt{20}\) is irrational.

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A proof of the irrationality of the square root of \(2\).

Proof by contradiction:

Assume that \(\sqrt{2}\) is rational.

Then by the definition of rational numbers, there exist two integers, a and \(b\) (where \(b \ne 0\), such that \(\sqrt{2}\ = \frac{a}{b}\).

The fraction \(\frac{a}{b}\) can be expressed in the irreducible form, \(\frac{c}{d}\), where \(c\) and \(d\) are relatively prime (ie. the greatest common divisor of both \(p\) and \(q\) is \(1\)), by dividing both \(a\) and \(b\) by their greatest common divisor: \(\sqrt{2} = \frac{a}{b} = \frac{\frac{a}{gcd(a,b)}}{\frac{b}{gcd(a,b)}} = \frac{c}{d}\)

Note that \(c\) and \(d\) are both still integers because, by definition of the greatest common divisor, \(gcd(a,b)\) divides both \(a\) and \(b\). Further, \(d \ne 0\) because \(b \ne 0\) and \(gcd(a,b) \geq 1\).

So, \(\sqrt{2}\ = \frac{c}{d}\), where \(c\) and \(d\) are integers, \(c\) and \(d\) are relatively prime, and \(d \ne 0\).

Squaring both sides, \(2 = (\frac{c}{d})^{2}\)

\(2 = \frac{c^{2}}{d^{2}}\)

\(2d^{2} = c^{2}\)

Thus, \(c^{2}\) must be divisible by \(2\). This is true only if \(c\) is divisible by \(2\) because the product of two even numbers is always even and the product of two odd numbers is always odd. So, \(c^{2}\) must be divisible by \(4\). There then exists some integer, \(q\), such that \(c^{2} = 4q\).

\(2d^{2} = 4q\)

\(d^{2} = 2q\)

So, \(d^{2}\) is divisible by \(2\), and thus \(d\) is divisible by \(2\).

Therefore, both \(c\) and \(d\) are divisible by \(2\). Because \(2\) is a common factor of both numbers, \(c\) and \(d\) are not relatively prime.

This is a contradiction because, as noted above, \(c\) and \(d\) are relatively prime.

Hence, \(\sqrt{2}\) cannot be rational. It is therefore irrational.

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The irrational numbers, while inexpressible as decimals, are still real numbers which are present in the world around us. What is the relationship between describability and validity?

A line having the length of an irrational number can be drawn with precision, but when written as a decimal the irrational will go on forever in a never-ending string of digits. Looking at the same object within a different context can expand the finite into something boundless or it can reduce the unexplainable into something tangible. How can those problems in your life which seem insurmountable be recontextualized into a manageable form? Can you take discrete moments of joy and expand them outward into something unlimited?

We can create the square roots of the positive integers through a single mechanism, and yet these generated roots differ in their fundamental natures. A given technique won't always produce the same results. Varying inputs, processed in the same way, can yield vastly differing outcomes. Explore the inputs that you're feeding into your own life's processes.

Rational numbers are those which can be represented as a fraction of two integers, having the form \(\frac{m}{n}\), where \(m\) and \(n\) are integers and \(n \ne 0\). Numbers such \(\frac{1}{2}\), \(\frac{3}{5}\), and \(\frac{14}{9}\) are rational numbers. The integers are also rational numbers, as they can be represented as themselves divided by \(1\) (eg. \(7 = \frac{7}{1}\)). The irrational numbers are numbers which cannot be represented this way. In their decimal form, they are represented by an infinite string of digits which do not exhibit any pattern. The most famous irrational number is perhaps \(\pi = 3.1415926…\), but others include Euler's number, \(e\), and the golden ratio, \(\phi\). Some of the square roots of numbers are irrational, including \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{20}\), \(\sqrt{130}\), and \(\sqrt{\frac{4}{5}}\), while other square roots of numbers are rational, such as \(\sqrt{4}\) = \(2\), \(\sqrt{9}\) = \(3\) \(\sqrt{144}\) = \(12\), \(\sqrt{\frac{9}{100}} = \frac{3}{10}\).

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A method for evaluating whether or not the square root of a number is irrational:

Reduce the number (or the numerator and the denominator of the number if the number is a fraction) under the square root into its prime factorization. If all of the prime factors have even exponents, then the square root of that number is rational. If one or more of the prime factors have odd exponents, then the square root of that number is irrational. Exploring \(\sqrt{144}\), the prime factorization of \(144\) is \(2^{4}3^{2}\). All of the prime factors, which in this case are \(2\) and \(3\), have even exponents, so \(\sqrt{144}\) is rational. Exploring \(\sqrt{20}\), the prime factorization of \(20\) is \(2^{2}5^{1}\). The prime factor \(5\) has an odd exponent, so \(\sqrt{20}\) is irrational.

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A proof of the irrationality of the square root of \(2\).

Proof by contradiction:

Assume that \(\sqrt{2}\) is rational.

Then by the definition of rational numbers, there exist two integers, a and \(b\) (where \(b \ne 0\), such that \(\sqrt{2}\ = \frac{a}{b}\).

The fraction \(\frac{a}{b}\) can be expressed in the irreducible form, \(\frac{c}{d}\), where \(c\) and \(d\) are relatively prime (ie. the greatest common divisor of both \(p\) and \(q\) is \(1\)), by dividing both \(a\) and \(b\) by their greatest common divisor: \(\sqrt{2} = \frac{a}{b} = \frac{\frac{a}{gcd(a,b)}}{\frac{b}{gcd(a,b)}} = \frac{c}{d}\)

Note that \(c\) and \(d\) are both still integers because, by definition of the greatest common divisor, \(gcd(a,b)\) divides both \(a\) and \(b\). Further, \(d \ne 0\) because \(b \ne 0\) and \(gcd(a,b) \geq 1\).

So, \(\sqrt{2}\ = \frac{c}{d}\), where \(c\) and \(d\) are integers, \(c\) and \(d\) are relatively prime, and \(d \ne 0\).

Squaring both sides, \(2 = (\frac{c}{d})^{2}\)

\(2 = \frac{c^{2}}{d^{2}}\)

\(2d^{2} = c^{2}\)

Thus, \(c^{2}\) must be divisible by \(2\). This is true only if \(c\) is divisible by \(2\) because the product of two even numbers is always even and the product of two odd numbers is always odd. So, \(c^{2}\) must be divisible by \(4\). There then exists some integer, \(q\), such that \(c^{2} = 4q\).

\(2d^{2} = 4q\)

\(d^{2} = 2q\)

So, \(d^{2}\) is divisible by \(2\), and thus \(d\) is divisible by \(2\).

Therefore, both \(c\) and \(d\) are divisible by \(2\). Because \(2\) is a common factor of both numbers, \(c\) and \(d\) are not relatively prime.

This is a contradiction because, as noted above, \(c\) and \(d\) are relatively prime.

Hence, \(\sqrt{2}\) cannot be rational. It is therefore irrational.

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The irrational numbers, while inexpressible as decimals, are still real numbers which are present in the world around us. What is the relationship between describability and validity?

A line having the length of an irrational number can be drawn with precision, but when written as a decimal the irrational will go on forever in a never-ending string of digits. Looking at the same object within a different context can expand the finite into something boundless or it can reduce the unexplainable into something tangible. How can those problems in your life which seem insurmountable be recontextualized into a manageable form? Can you take discrete moments of joy and expand them outward into something unlimited?

We can create the square roots of the positive integers through a single mechanism, and yet these generated roots differ in their fundamental natures. A given technique won't always produce the same results. Varying inputs, processed in the same way, can yield vastly differing outcomes. Explore the inputs that you're feeding into your own life's processes.