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Rational self-interest implies that

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โˆ™ 2009-07-31 21:26:40
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Show that the sum of rational no with an irrational no is always 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.


Prove R is rational and S is irrational then R plus S must be irrational?

Let R + S = T, and suppose that T is a rational number.The set of rational number is a group.This implies that since R is rational, -R is rational [invertibility].Then, since T and -R are rational, T - R must be rational [closure].But T - R = S which implies that S is rational.That contradicts the fact that y is an irrational number. The contradiction implies that the assumption [that T is rational] is incorrect.Thus, the sum of a rational number R and an irrational number S cannot be rational.


If you add a rational and irrational number what is the sum?

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.


What is another example of the five steps in solving rational equation?

"another" implies that you already have one example. In order to answer the question it might just help to know what that is.


Explain why the sum of a rational number and an irrational number is an irrational number?

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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How do you show that 2 plus root2 is an irrational number?

Assume it's rational. Then 2 + root2 = some rational number q. Then root2 = q - 2. However, the rational numbers are well-defined under addition by (a,b) + (c,d) = (ad + bc, bd) (in other words, you can add two fractions a/b and c/d and always get another fraction of the form (ad + bc)/bd.) Therefore, q - 2 = q + (-2) is rational, since both q and -2 are rational. This implies root2 must be rational, which is a contradiction. Therefore the assumption that 2 + root2 is rational must be false.


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