Let R1 = rational number
Let X = irrational number
Assume R1 + X = (some rational number) We add -R1 to both sides, and we get: -R1 + x = (some irrational number) + (-R1), thus X = (SIR) + (-R1), which implies that X, an irrational number, is the sum of two rational numbers, which is a contradiction. Thus, the sum of a rational number and an irrational number is always irrational. (Proof by contradiction)
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No. The sum of an irrational number and any other [real] number is irrational.
an irrational number PROOF : Let x be any rational number and y be any irrational number. let us assume that their sum is rational which is ( z ) x + y = z if x is a rational number then ( -x ) will also be a rational number. Therefore, x + y + (-x) = a rational number this implies that y is also rational BUT HERE IS THE CONTRADICTION as we assumed y an irrational number. Hence, our assumption is wrong. This states that x + y is not rational. HENCE PROVEDit will always be irrational.
Suppose x is a rational number and y is an irrational number.Let x + y = z, and assume that z is a rational number.The set of rational number is a group.This implies that since x is rational, -x is rational [invertibility].Then, since z and -x are rational, z - x must be rational [closure].But z - x = y which implies that y is rational.That contradicts the fact that y is an irrational number. The contradiction implies that the assumption [that z is rational] is incorrect.Thus, the sum of a rational number x and an irrational number y cannot be rational.
The sum or the difference between two irrational numbers could either be rational or irrational, however, it should be a real number.