Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Proof that the square root of any nonperfect square positive integer is irrational

Revision as of 12:50, 15 June 2019 by Colball (talk | contribs)

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.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Proof_that_the_square_root_of_any_nonperfect_square_positive_integer_is_irrational&oldid=106350"