1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Prove that $\frac{\left(3-4\sqrt{2}\right)}{7}$ is an irrational number, given that $\sqrt{2}$ is an irrational number.

Open in App
Solution

## Let us assume that $\frac{\left(3-4\sqrt{2}\right)}{7}$ is a rational number. Thus, $\frac{\left(3-4\sqrt{2}\right)}{7}$ can be represented in the form of $\frac{p}{q},$ where p and q are integers, q ≠ 0, p and q are co-prime numbers. $\frac{3-4\sqrt{2}}{7}=\frac{p}{q}\phantom{\rule{0ex}{0ex}}⇒3-4\sqrt{2}=\frac{7p}{q}\phantom{\rule{0ex}{0ex}}⇒4\sqrt{2}=3-\frac{7p}{q}\phantom{\rule{0ex}{0ex}}⇒4\sqrt{2}=\frac{3q-7p}{q}\phantom{\rule{0ex}{0ex}}⇒\sqrt{2}=\frac{3q-7p}{4q}\phantom{\rule{0ex}{0ex}}\mathrm{Since},\frac{3q-7p}{4q}\mathrm{is}\mathrm{rational}⇒\sqrt{2}\mathrm{is}\mathrm{rational}\phantom{\rule{0ex}{0ex}}\mathrm{But},\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}\sqrt{2}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{our}\mathrm{assumption}\mathrm{is}\mathrm{wrong}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\frac{3-4\sqrt{2}}{7}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}.$

Suggest Corrections
62
Join BYJU'S Learning Program
Related Videos
Revisiting Irrational Numbers
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program