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Why does the sum of rational number and irrational numbers are always irrational?
Let your sum be a + b = c, where "a" is irrational, "b" is rational, and "c" may be either (that's what we want to find out). In this case, c - b = a. If we assume that c is rational, you would have: a rational number minus a rational number is an irrational number, which can't be true (both addition and subtraction are closed in the set of rational numbers). Therefore, we have a contradiction with the assumption that "c" (the sum in the original equation) is rational.
What makes a number rational?
A rational number is one that can be expressed as ratio(or fraction) of two integers: given integers a & b (with b ≠ 0), then the fraction a / b (or a ÷ b) will be a rational number. We need to specify that b not equal to zero, as division by zero is not defined (well not for most math applications).So irrational numbers such as pi, e, and square root of 2: cannot in any way be resolved into a fraction of two integers. All rational numbers can. Examples are:0 is rational, let a = 0, b = 1, or any other non-zero integer: 0/1 = 0Whole numbers are rational: 1/1 = 1; 2/1 = 2; 100/25 = 4, etc.Negative integers are also rational: (-4)/(2) = -2, or (20)/(-5) = -4All fractions (proper and improper) which have integers in the numerator, and non-zero integer in denominator, as well as mixed numbers are rational {positive and negative} will be rational numbers.
What operation is the set of positive rational numbers not closed?
subtraction. Let's take 1/2 and subtract 3/4 which is great than 1/2 so the answer is negative and hence not a positive rational.
Why is the sum of a rational numbers and an irrational number is irrational?
Let x be a rational number and y be an irrational number.Suppose their sum = z, is rational.That is x + y = zThen y = z - xThe set of rational number is closed under addition (and subtraction). Therefore, z - x is rational.Thus you have left hand side (irrational) = right hand side (rational) which is a contradiction.Therefore, by reducio ad absurdum, the supposition that z is rational is false, ie the sum of a rational and an irrational must be irrational.
Is square root of 36 rational?
Hmm, let me see ... 6. Since I'm not still writing decimals, yes, it's rational.
