Rational and irrational numbers/Pythagorean triple in unit interval/Exercise/Solution

a) We have

${}3^{2}+4^{2}=5^{2}\,,$

hence

${}{\left({\frac {3}{6}}\right)}^{2}+{\left({\frac {4}{6}}\right)}^{2}={\left({\frac {5}{6}}\right)}^{2}\,,$

and these numbers are rational and within the open unit interval.

b) We take ${}a={\frac {3}{6}}$, ${}b={\frac {4}{6}}$ and ${}c={\frac {1}{6}}$. The sum is

${}a^{2}+b^{2}={\left({\frac {5}{6}}\right)}^{2}\neq {\left({\frac {1}{6}}\right)}^{2}\,.$

c) We put

${}a=b={\frac {1}{2{\sqrt {2}}}}\,;$

this number is irrational, because ${}{\sqrt {2}}$ is irrational. We have

${}{\left({\frac {1}{2{\sqrt {2}}}}\right)}^{2}+{\left({\frac {1}{2{\sqrt {2}}}}\right)}^{2}={\frac {1}{4\cdot 2}}+{\frac {1}{4\cdot 2}}={\frac {1}{4}}={\left({\frac {1}{2}}\right)}^{2}\,.$

With ${}c={\frac {1}{2}}$ we have found an example.

To the exercise