Question

prove that 2+√3 is irrational​

Answer 1

Answer:

let 2+√3 = rational

then, 2+√3 = p/q

√3 = p/q  -  2

√3 = (p-2q)/q         (taking lcm)

since √3 is irrational and (p-2q)/q is rational,

it means that irrational = rational,

but this contradicts the fact that √3 is irrational.

hence, 2+√3 is irrational

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Answer 2

We assume that 2 + √3 is a rational number.

We assume that 2 + √3 is a rational number.=> 2 + √3 = p/q , where p & q are integers, ‘q’ not = 0.

We assume that 2 + √3 is a rational number.=> 2 + √3 = p/q , where p & q are integers, ‘q’ not = 0.=> √3 = (p/q) - 2

We assume that 2 + √3 is a rational number.=> 2 + √3 = p/q , where p & q are integers, ‘q’ not = 0.=> √3 = (p/q) - 2=> √3 = (p - 2q)/ q ………… (1)

We assume that 2 + √3 is a rational number.=> 2 + √3 = p/q , where p & q are integers, ‘q’ not = 0.=> √3 = (p/q) - 2=> √3 = (p - 2q)/ q ………… (1)=> here, LHS √3 is an irrational number.

We assume that 2 + √3 is a rational number.=> 2 + √3 = p/q , where p & q are integers, ‘q’ not = 0.=> √3 = (p/q) - 2=> √3 = (p - 2q)/ q ………… (1)=> here, LHS √3 is an irrational number.But RHS is a rational number.. Reason- the difference of 2 integers is always an integer. So the numerator (p- 2q) is an integer.

We assume that 2 + √3 is a rational number.=> 2 + √3 = p/q , where p & q are integers, ‘q’ not = 0.=> √3 = (p/q) - 2=> √3 = (p - 2q)/ q ………… (1)=> here, LHS √3 is an irrational number.But RHS is a rational number.. Reason- the difference of 2 integers is always an integer. So the numerator (p- 2q) is an integer.& the denominator ‘q’ is an integer.&‘q’ not = 0

We assume that 2 + √3 is a rational number.=> 2 + √3 = p/q , where p & q are integers, ‘q’ not = 0.=> √3 = (p/q) - 2=> √3 = (p - 2q)/ q ………… (1)=> here, LHS √3 is an irrational number.But RHS is a rational number.. Reason- the difference of 2 integers is always an integer. So the numerator (p- 2q) is an integer.& the denominator ‘q’ is an integer.&‘q’ not = 0This way, all conditions of a rational number are satisfied.

We assume that 2 + √3 is a rational number.=> 2 + √3 = p/q , where p & q are integers, ‘q’ not = 0.=> √3 = (p/q) - 2=> √3 = (p - 2q)/ q ………… (1)=> here, LHS √3 is an irrational number.But RHS is a rational number.. Reason- the difference of 2 integers is always an integer. So the numerator (p- 2q) is an integer.& the denominator ‘q’ is an integer.&‘q’ not = 0This way, all conditions of a rational number are satisfied.=> RHS (p- 2q)/q is a rational number.

We assume that 2 + √3 is a rational number.=> 2 + √3 = p/q , where p & q are integers, ‘q’ not = 0.=> √3 = (p/q) - 2=> √3 = (p - 2q)/ q ………… (1)=> here, LHS √3 is an irrational number.But RHS is a rational number.. Reason- the difference of 2 integers is always an integer. So the numerator (p- 2q) is an integer.& the denominator ‘q’ is an integer.&‘q’ not = 0This way, all conditions of a rational number are satisfied.=> RHS (p- 2q)/q is a rational number.But , LHS is an irrational.

We assume that 2 + √3 is a rational number.=> 2 + √3 = p/q , where p & q are integers, ‘q’ not = 0.=> √3 = (p/q) - 2=> √3 = (p - 2q)/ q ………… (1)=> here, LHS √3 is an irrational number.But RHS is a rational number.. Reason- the difference of 2 integers is always an integer. So the numerator (p- 2q) is an integer.& the denominator ‘q’ is an integer.&‘q’ not = 0This way, all conditions of a rational number are satisfied.=> RHS (p- 2q)/q is a rational number.But , LHS is an irrational.=> LHS of….. (1) is not = RHS.

We assume that 2 + √3 is a rational number.=> 2 + √3 = p/q , where p & q are integers, ‘q’ not = 0.=> √3 = (p/q) - 2=> √3 = (p - 2q)/ q ………… (1)=> here, LHS √3 is an irrational number.But RHS is a rational number.. Reason- the difference of 2 integers is always an integer. So the numerator (p- 2q) is an integer.& the denominator ‘q’ is an integer.&‘q’ not = 0This way, all conditions of a rational number are satisfied.=> RHS (p- 2q)/q is a rational number.But , LHS is an irrational.=> LHS of….. (1) is not = RHS.=> Our assumption, that 2 + √3 is a rational number, is incorrect..

We assume that 2 + √3 is a rational number.=> 2 + √3 = p/q , where p & q are integers, ‘q’ not = 0.=> √3 = (p/q) - 2=> √3 = (p - 2q)/ q ………… (1)=> here, LHS √3 is an irrational number.But RHS is a rational number.. Reason- the difference of 2 integers is always an integer. So the numerator (p- 2q) is an integer.& the denominator ‘q’ is an integer.&‘q’ not = 0This way, all conditions of a rational number are satisfied.=> RHS (p- 2q)/q is a rational number.But , LHS is an irrational.=> LHS of….. (1) is not = RHS.=> Our assumption, that 2 + √3 is a rational number, is incorrect..=> 2 + √3 is an irrational number

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