Let assume that 7-2root3 be rational and rational numbers are in the form of p/q form.We know that root 3 is irrational and not p/q form. It contradicts the statement that it is irrational as it is p/q form.Subtraction and multiplication of irrational number form irrational number so\xa07-2root3 is irrational.

\xa0let us assume that\xa0{tex}\\sqrt 3{/tex}\xa0be a rational number.{tex}\\sqrt { 3 } = \\frac { a } { b }{/tex}, where a and b are integers and co-primes and b{tex} \\neq{/tex}0Squaring both sides, we have{tex}\\frac { a ^ { 2 } } { b ^ { 2 } } = 3{/tex}or,\xa0{tex}a ^ { 2 } = 3 b ^ { 2 }{/tex}——–(i)a2\xa0is 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\xa0= 3b29c2\xa0= 3b2b2\xa0= 3c2so b2\xa0is divisible by 3hence, b is divisible by 3……….(iii)From equation(ii) and (iii), we have3 is a factor of a and b which is contradicting the fact that a and b are co-primes.Thus, our assumption that\xa0{tex}\\sqrt 3{/tex} is rational number is wrong.Hence,\xa0{tex}\\sqrt 3{/tex}\xa0is an irrational number.Let us assume that 7 – 2{tex}\\sqrt 3{/tex}\xa0is a rational number7 -\xa02{tex}\\sqrt 3{/tex}\xa0=\xa0{tex}\\frac { p } { q }{/tex} (q\xa0{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\xa0and 2q both are integers, hence\xa0{tex}\\sqrt 3{/tex} is a rational number.But this contradicts the fact that\xa0{tex}\\sqrt 3{/tex}\xa0is an irrational number.This contradict is due to our assumption that\xa0{tex}\\style{font-family:Arial}{7-2\\sqrt3}{/tex}\xa0is rational.Hence, 7 -\xa02{tex}\\sqrt3{/tex}\xa0is irrational.

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Question:- (7-2√3)Solution:- Let us assume that (7-2√3) is a rational number.Therefore we can write in the form of p/q. Where p and q are co- prime numbers.7-2√3=p/q7-p/q=2√37q-p/q=2√37q-p/2q=√3√3=7q-p/2qSince, we know that √3 is an irrational number .Hence our assumption is wrong 7q-p/q or 7-2√3 is an irrational number.. Hence Proved…..I hope now your doubt is clear??