Prove that 7-2*3 is an irrational …

 let us assume that {tex}\sqrt 3{/tex} be a rational number.
{tex}\sqrt { 3 } = \frac { a } { b }{/tex}, where a and b are integers and co-primes and b{tex} \neq{/tex}0
Squaring both sides, we have
{tex}\frac { a ^ { 2 } } { b ^ { 2 } } = 3{/tex}
or, {tex}a ^ { 2 } = 3 b ^ { 2 }{/tex}--------(i)
a² is divisible by 3.
Hence a is divisible by 3..........(ii)
Let a = 3c ( where c is any integer)

squaring on both sides we get
(3c)² = 3b²
9c² = 3b²
b² = 3c²
so b² is divisible by 3
hence, b is divisible by 3..........(iii)
From equation(ii) and (iii), we have
3 is a factor of a and b which is contradicting the fact that a and b are co-primes.
Thus, our assumption that {tex}\sqrt 3{/tex} is rational number is wrong.
Hence, {tex}\sqrt 3{/tex} is an irrational number.
Let us assume that 7 - 2{tex}\sqrt 3{/tex} is a rational number
7 - 2{tex}\sqrt 3{/tex} = {tex}\frac { p } { q }{/tex}   (q {tex}\neq{/tex}0 and p and q are co-primes)
{tex}\style{font-family:Arial}{\begin{array}{l}7-2\sqrt3=\frac pq\\-2\sqrt3=\frac pq-7\\2\sqrt3=7-\frac pq\\2\sqrt3=\frac{7q-p}q\\\sqrt3=\frac{7q-p}{2q}\end{array}}{/tex}
Here 7q-p and 2q both are integers, hence {tex}\sqrt 3{/tex} is a rational number.
But this contradicts the fact that {tex}\sqrt 3{/tex} is an irrational number.
This contradict is due to our assumption that {tex}\style{font-family:Arial}{7-2\sqrt3}{/tex} is rational.
Hence, 7 - 2{tex}\sqrt3{/tex} is irrational.