Question

Verify by the method of contradiction that $\sqrt{7}$ is irrational.

Answer 1

Hint: Here, we have to prove that $\sqrt{7}$ is irrational by the method of contradiction. For this, we will first assume that the given statement is false, i.e. $\sqrt{7}$ is rational. Then we will write it in the form of $\dfrac{p}{q}$ where p and q are co prime integers. Then we will try to prove that 7 is a common factor to both p and q by squaring the thus formed equations and forming more equations of the same format. This will prove that our assumption is wrong as we assumed that p and q are co prime integers. Hence, we will prove the given statement.

Complete step-by-step answer:
To prove that $\sqrt{7}$ is irrational, we will assume that the given statement is false, i.e. we will assume that $\sqrt{7}$ is rational.
Now, we know that any rational number can be written in the form of $\dfrac{p}{q}$ where p and q are co-prime integers.
Thus, we can say that:
$\sqrt{7}=\dfrac{p}{q}$
Now, squaring on both sides, we get:
$\begin{align}
  & {{\left( \sqrt{7} \right)}^{2}}={{\left( \dfrac{p}{q} \right)}^{2}} \\
 & \Rightarrow 7=\dfrac{{{p}^{2}}}{{{q}^{2}}} \\
 & \Rightarrow {{p}^{2}}=7{{q}^{2}} \\
\end{align}$
$\Rightarrow {{q}^{2}}=\dfrac{{{p}^{2}}}{7}$ …..(i)
Now, since, as mentioned above, p and q are integers, we can say from equation (i) that ${{p}^{2}}$ is divisible by 7.
Now, since ${{p}^{2}}$ is divisible by 7, p is also divisible by 7.
Thus, we can say that 7 is a factor of p.
Now, since 7 is a factor of p, we can write p as a multiple of 7 and m where m is any integer.
Thus, we can say that:
$p=7m$
Now, squaring both sides we get:
$\begin{align}
  & {{\left( p \right)}^{2}}={{\left( 7m \right)}^{2}} \\
 & \Rightarrow {{p}^{2}}=49{{m}^{2}} \\
\end{align}$
Now, putting the value of ${{p}^{2}}$ as $7{{q}^{2}}$ from the equation obtained above we get:
$\begin{align}
  & {{p}^{2}}=49{{m}^{2}} \\
 & \Rightarrow 7{{q}^{2}}=49{{m}^{2}} \\
 & \Rightarrow {{q}^{2}}=7{{m}^{2}} \\
\end{align}$
$\Rightarrow {{m}^{2}}=\dfrac{{{q}^{2}}}{7}$ …..(ii)
Now, since m is an integer, ${{m}^{2}}$ will also be an integer. Hence, $\dfrac{{{q}^{2}}}{7}$ is also an integer which implies that ${{q}^{2}}$ is divisible by 7.
Now, since ${{q}^{2}}$ is divisible by 7, q is also divisible by 7.
Hence, we can say that 7 is a factor of q.
Now, we mentioned above that p and q are co-prime numbers but here we have proved that 7 is a factor common to both p and q.
Hence, the assumption we made, i.e. $\sqrt{7}$ is rational, is wrong.
Thus, we can say that $\sqrt{7}$ is irrational.
Hence, proved.

Note: These kind of questions are very important and it may be given not only for $\sqrt{7}$ but also for other irrational numbers like $\sqrt{2},\sqrt{3}$ and $\sqrt{5}$. They will be proved irrational in the exact same way as we proved $\sqrt{7}$ above. Then, we may also get numbers like $\sqrt{7}+3,\sqrt{3}-2$ , etc. These may be proved by first separating the rational and irrational quantities of either side of the equal to sign by assuming it equal to $\dfrac{p}{q}$ and then stating then proving the irrational quantities rational in the same way as above and then stating that rational and irrational quantities cannot be equal.