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How do you find irrational numbers between two rational numbers? - Answers

If a and b are rational, with a < b, then a + (b-a) [sqrt(2)/ 2] is an irrational number between a and b. This number is between a and b because sqrt(2)/2 is less than one and positive, so that a < a + (b-a) [sqrt(2)/3] < a + (b-a) [1] = b. To prove that a + (b-a) [sqrt(2)/2] is not rational, suppose that a + (b-a) [sqrt(2)/2] = p/q where p and q are integers. Then, sqrt(2) = ( p/q -a ) 2/(b-a) which is rational since the rationals are a field, closed under arithmetical operation, but sqrt(2) not rational (Look up the elementary proof if you do not know it.)



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How do you find irrational numbers between two rational numbers? - Answers

https://math.answers.com/basic-math/How_do_you_find_irrational_numbers_between_two_rational_numbers

If a and b are rational, with a < b, then a + (b-a) [sqrt(2)/ 2] is an irrational number between a and b. This number is between a and b because sqrt(2)/2 is less than one and positive, so that a < a + (b-a) [sqrt(2)/3] < a + (b-a) [1] = b. To prove that a + (b-a) [sqrt(2)/2] is not rational, suppose that a + (b-a) [sqrt(2)/2] = p/q where p and q are integers. Then, sqrt(2) = ( p/q -a ) 2/(b-a) which is rational since the rationals are a field, closed under arithmetical operation, but sqrt(2) not rational (Look up the elementary proof if you do not know it.)



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https://math.answers.com/basic-math/How_do_you_find_irrational_numbers_between_two_rational_numbers

How do you find irrational numbers between two rational numbers? - Answers

If a and b are rational, with a < b, then a + (b-a) [sqrt(2)/ 2] is an irrational number between a and b. This number is between a and b because sqrt(2)/2 is less than one and positive, so that a < a + (b-a) [sqrt(2)/3] < a + (b-a) [1] = b. To prove that a + (b-a) [sqrt(2)/2] is not rational, suppose that a + (b-a) [sqrt(2)/2] = p/q where p and q are integers. Then, sqrt(2) = ( p/q -a ) 2/(b-a) which is rational since the rationals are a field, closed under arithmetical operation, but sqrt(2) not rational (Look up the elementary proof if you do not know it.)

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      If a and b are rational, with a < b, then a + (b-a) [sqrt(2)/ 2] is an irrational number between a and b. This number is between a and b because sqrt(2)/2 is less than one and positive, so that a < a + (b-a) [sqrt(2)/3] < a + (b-a) [1] = b. To prove that a + (b-a) [sqrt(2)/2] is not rational, suppose that a + (b-a) [sqrt(2)/2] = p/q where p and q are integers. Then, sqrt(2) = ( p/q -a ) 2/(b-a) which is rational since the rationals are a field, closed under arithmetical operation, but sqrt(2) not rational (Look up the elementary proof if you do not know it.)

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