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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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Q: How do you find irrational numbers between two rational numbers?
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A rational number is one that is the ratio of two integers, like 3/4 or 355/113. An irrational number can't be expressed as the ratio of any two integers, and examples are the square root of 2, and pi. Between any two rational numbers there is an irrational number, and between any two irrational numbers there is a rational number.


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An irrational number is expressed as a non-repeating decimal that goes on forever. Write out the enough of the decimal expansion of each number to find the first digit where the two numbers disagree. Truncate the larger number at that digit, and the result is a rational number (terminating decimal) that is between the two.


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Why does the sum of rational number and irrational numbers are always irrational?

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How can rational number be used to help locate irrational?

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