Question

Prove the following are irrational numbers,
$13+25\sqrt{2}$

Answer 1

Hint: We will use the contradiction method to prove it. We will first start by letting the number to be a rational one and representing it in the form of $\dfrac{a}{b}$ and then use it to prove that $\sqrt{2}$ is rational which contradicts the fact that $\sqrt{2}$ is irrational.

Complete step-by-step answer:
Now, we have to prove that $13+25\sqrt{2}$ is irrational.
We will the contradiction of that $13+25\sqrt{2}$ is irrational number and let that $13+25\sqrt{2}$ is rational.
Now, we know that a rational number can be represented as $\dfrac{a}{b}$ where a and b are co – prime and $b\ne 0$.
So, we have,
$\begin{align}
  & 13+25\sqrt{2}=\dfrac{a}{b} \\
 & 25\sqrt{2}=\dfrac{a}{b}-13 \\
 & 25\sqrt{2}=\dfrac{a-13b}{b} \\
 & \sqrt{2}=\dfrac{a-13b}{25b} \\
\end{align}$
Now, we know that,
$\begin{align}
  & Integer+Integer=Integer \\
 & Integer-Integer=Integer \\
 & Integer\times Integer=Integer \\
 & \dfrac{Integer}{Integer}=Integer \\
\end{align}$
Therefore, we have $\dfrac{a-13b}{25b}$ as rational and hence, $\sqrt{2}$ as a rational number but this contradict the fact that $\sqrt{2}$ is irrational. Hence, our assumption is wrong and $13+25\sqrt{2}$ is an irrational number.

Note: It is important to note that a rational number can be represented as $\dfrac{p}{q}$ where p, q are co – primes and $q\ne 0$. Also, it is important that if we perform any operation like addition, subtraction then the resultant is also an integer.