Remove the irrationality from the denominator 1/1+root2+root3
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We want a rational number a/b such that:
sqrt(2) < a/b < sqrt(3)
Squaring each term above gets:
2 < a^2/b^2 < 3
Looking at the set of squared integers {0, 1, 4, 9, 16, 25, ...} it's easy to find a pair such that their quotient is greater than 2 but less than 3 -- for example, 9/4:
2 < 9/4 < 3 , so sqrt(2) < 3/2 < sqrt(3)
Note that there is an unlimited supply of others that work, too:
2 < 25/9 < 3 , so sqrt(2) < 5/3 < sqrt(3)
2 < 64/25 < 3 , so sqrt(2) < 8/5 < sqrt(3)
2 < 121/49 < 3 , so sqrt(2) < 11/7 < sqrt(3)
etcetera.
Anonymous · 1 decade ago
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We want a rational number a/b such that:
sqrt(2) < a/b < sqrt(3)
Squaring each term above gets:
2 < a^2/b^2 < 3
Looking at the set of squared integers {0, 1, 4, 9, 16, 25, ...} it's easy to find a pair such that their quotient is greater than 2 but less than 3 -- for example, 9/4:
2 < 9/4 < 3 , so sqrt(2) < 3/2 < sqrt(3)
Note that there is an unlimited supply of others that work, too:
2 < 25/9 < 3 , so sqrt(2) < 5/3 < sqrt(3)
2 < 64/25 < 3 , so sqrt(2) < 8/5 < sqrt(3)
2 < 121/49 < 3 , so sqrt(2) < 11/7 < sqrt(3)
etcetera.
Anonymous · 1 decade ago
Comment
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