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The Irrationality ofProblem:
Solution:
for a and b = any two integers. To show that is irrational, we must show that no two such integers can be found. We begin by squaring both sides of eq. 1: From eq. 2a, we must conclude that a^{2} (and, therefore, a) is even; b^{2} (and, therefore, b) may be even or odd. If b is even, the ratio a^{2}/b^{2} may be immediately reduced by canceling a common factor of 2. If b is odd, it is possible that the ratio a^{2}/b^{2} is already reduced to smallest possible terms. We assume that b^{2} (and, therefore, b) is odd. Now, we set a = 2m, and b = 2n + 1, and require that m and n be integers (to ensure integer values of a and b). Then Substituting these expressions into eq. 2a, we obtain The L.H.S. of eq. 6 is an odd integer. The R.H.S., on the other hand, is an even integer. There are no solutions for eq. 6. Therefore, integer values of a and b which satisfy the relationship = a/b cannot be found. We are forced to conclude that is irrational. |
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