Question

prove that (2-7√3) is an irrational number where √3 is irrational​

Answer 1

Step-by-step explanation:

Let us assume in contradiction that 2√3-7 is rational number. Let 2√3-7 =a/b where a and b are co-primes. ... 2√3-7 is irrational number

Answer 2

Step-by-step explanation:

Solution:

If possible, let $2 -7 \sqrt{3}$ be rational and let its simplest form be $\frac{a}{b}$,where a and b are integers and b≠0

Then,

⇛ $2 -7 \sqrt{3}$ = $\frac{a}{b}$

⇛ $-7 \sqrt{3}$ = $\frac{a}{b}$ - 2

⇛ $\sqrt{3}$ = $\frac{-1}{7}$($\frac{a}{b}$ - 2)

⇛ $\sqrt{3}$ = ($\frac{ - a}{7b} + \frac{2}{7}$)... i

Now, $\frac{a}{b}$ is rational

⇛ $\frac{-a}{7b}$ is rational

⇛ ($\frac{ - a}{7b} + \frac{2}{7}$) is rational (∵ difference of rationals is rational)

⇛ $\sqrt{3}$ is rational [from (i)]

• But, it is given that $\sqrt{3}$ is irrational.
• Thus, we arrive at a contradiction.
• Since the contradiction arises by assuming that ($2 -7 \sqrt{3}$) is rational
• Hence, ($2 -7 \sqrt{3}$) is irrational.