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Difference between revisions of "Proof that the square root of any nonperfect square positive integer is irrational"
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PROOF 1:Let us assume that <math>\sqrt{n}</math> is rational where <math>n</math> is a nonperfect square positive integer. Then it can be written as <math>\frac{p}{q}=n^2 \Rightarrow \frac{p^2}{q^2}=n \Rightarrow (q^2)n=p^2</math>. But no perfect square times a nonperfect square positive integer is a perfect square. Therefore <math>\sqrt{n}</math> is irrational.
 
PROOF 1:Let us assume that <math>\sqrt{n}</math> is rational where <math>n</math> is a nonperfect square positive integer. Then it can be written as <math>\frac{p}{q}=n^2 \Rightarrow \frac{p^2}{q^2}=n \Rightarrow (q^2)n=p^2</math>. But no perfect square times a nonperfect square positive integer is a perfect square. Therefore <math>\sqrt{n}</math> is irrational.
  
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Proof 2:Suppose to the contrary that <math>\sqrt{2} = a/b</math> for integers a,b, and that this representation is fully reduced, so that <math>\textup{gcd}(a,b) = 1</math>. Consider the isosceles right triangle with side length b and hypotenuse length a, as in the picture on the left. Indeed, by the Pythagorean theorem, the length of the hypotenuse is <math>\sqrt{b^2 + b^2} = b \sqrt{2} = a</math>, since <math>\sqrt{2} = a/b</math>.
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Proof 2:Suppose to the contrary that \sqrt{2} = a/b for integers a,b, and that this representation is fully reduced, so that \textup{gcd}(a,b) = 1. Consider the isosceles right triangle with side length b and hypotenuse length a, as in the picture on the left. Indeed, by the Pythagorean theorem, the length of the hypotenuse is \sqrt{b^2 + b^2} = b \sqrt{2} = a, since \sqrt{2} = a/b.
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Swinging a b-leg to the hypotenuse, as shown, we see that the hypotenuse can be split into parts b, a-b, and hence a-b is an integer. Call the point where the b and a-b parts meet P. If we extend a perpendicular line from P to the other leg, as shown, we get a second, smaller isosceles right triangle. Since the segments PQ and QR are symmetrically aligned (they are tangents to the same circle from the same point), they too have a length equal to a-b. Finally, we may write the hypotenuse of the smaller triangle as <math>b-(a-b) = 2b-a</math>, which is also an integer. Should the lengths of the sides of the smaller triangle are integers, but by triangle similarity, the hypotenuse to side-length ratios are equal: <math>\sqrt{2} = a/b = (2b-a)/(a-b)</math>, and obviously from the picture the latter numerator and denominator are smaller numbers. Hence, <math>a/b</math> was not in lowest terms, a contradiction. This implies that <math>\sqrt{2}</math> cannot be rational.
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Swinging a b-leg to the hypotenuse, as shown, we see that the hypotenuse can be split into parts b, a-b, and hence a-b is an integer. Call the point where the b and a-b parts meet P. If we extend a perpendicular line from P to the other leg, as shown, we get a second, smaller isosceles right triangle. Since the segments PQ and QR are symmetrically aligned (they are tangents to the same circle from the same point), they too have a length equal to a-b. Finally, we may write the hypotenuse of the smaller triangle as b-(a-b) = 2b-a, which is also an integer. Should the lengths of the sides of the smaller triangle are integers, but by triangle similarity, the hypotenuse to side-length ratios are equal: \sqrt{2} = a/b = (2b-a)/(a-b), and obviously from the picture the latter numerator and denominator are smaller numbers. Hence, a/b was not in lowest terms, a contradiction. This implies that \sqrt{2} cannot be rational. \square

Revision as of 12:50, 15 June 2019

PROOFS

PROOF 1:Let us assume that $\sqrt{n}$ is rational where $n$ is a nonperfect square positive integer. Then it can be written as $\frac{p}{q}=n^2 \Rightarrow \frac{p^2}{q^2}=n \Rightarrow (q^2)n=p^2$. But no perfect square times a nonperfect square positive integer is a perfect square. Therefore $\sqrt{n}$ is irrational.

Proof 2:Suppose to the contrary that \sqrt{2} = a/b for integers a,b, and that this representation is fully reduced, so that \textup{gcd}(a,b) = 1. Consider the isosceles right triangle with side length b and hypotenuse length a, as in the picture on the left. Indeed, by the Pythagorean theorem, the length of the hypotenuse is \sqrt{b^2 + b^2} = b \sqrt{2} = a, since \sqrt{2} = a/b. Swinging a b-leg to the hypotenuse, as shown, we see that the hypotenuse can be split into parts b, a-b, and hence a-b is an integer. Call the point where the b and a-b parts meet P. If we extend a perpendicular line from P to the other leg, as shown, we get a second, smaller isosceles right triangle. Since the segments PQ and QR are symmetrically aligned (they are tangents to the same circle from the same point), they too have a length equal to a-b. Finally, we may write the hypotenuse of the smaller triangle as b-(a-b) = 2b-a, which is also an integer. Should the lengths of the sides of the smaller triangle are integers, but by triangle similarity, the hypotenuse to side-length ratios are equal: \sqrt{2} = a/b = (2b-a)/(a-b), and obviously from the picture the latter numerator and denominator are smaller numbers. Hence, a/b was not in lowest terms, a contradiction. This implies that \sqrt{2} cannot be rational. \square

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