Q: How do you show irrational number r uncountable?

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No: Let r be some irrational number; as such it cannot be represented as s/t where s and t are both non-zero integers. Assume the square root of this irrational number r was rational. Then it can be represented in the form of p/q where p and q are both non-zero integers, ie √r = p/q As p is an integer, p² = p×p is also an integer, let y = p² And as q is an integer, q² = q×q is also an integer, let x = q² The number is the square of its square root, thus: (√r)² = (p/q)² = p²/q² = y/x but (√r)² = r, thus r = y/x and is a rational number. But r was chosen to be an irrational number, which is a contradiction (r cannot be both rational and irrational at the same time, so it cannot exist). Thus the square root of an irrational number cannot be rational. However, the square root of a rational number can be irrational, eg for the rational number ½ its square root (√½) is not rational.

Yes. The proof is easy. Let x be the irrational number and assume there exists some rational number r = a/b where a and b are integers (that's what it means to be rational). Now suppose x/r is a rational number. Then x/r = (b/a)x = c/d where c and d are some other integers. Since (b/a)x=c/d, then x = bd/ac which means that x itself is rational, but we assumed it was irrational. The contradiction proves that the assertion is wrong. An irrational divided by a rational must be irrational.

The set of irrational numbers is NOT denoted by Q.Q denotes the set of rational numbers. The set of irrational numbers is not denoted by any particular letter but by R - Q where R is the set of real numbers.

There is no specific sign. The set of irrationals can be written as R - Q.

The radius, r, is 4.8 feet. The circumference is 2*pi*r where pi is the irrational number approximated by 3.14159

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Proof By Contradiction:Claim: R\Q = Set of irrationals is countable.Then R = Q union (R\Q)Since Q is countable, and R\Q is countable (by claim), R is countable because the union of countable sets is countable.But this is a contradiction since R is uncountable (Cantor's Diagonal Argument).Thus, R\Q is uncountable.

It the radius is r then the area is pi*r*r - which is pi times a rational number. pi is an irrational number, so the multiple of pi and a rational number is irrational.

One way is Cantor's diagonal argument. See link.

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.

No: Let r be some irrational number; as such it cannot be represented as s/t where s and t are both non-zero integers. Assume the square root of this irrational number r was rational. Then it can be represented in the form of p/q where p and q are both non-zero integers, ie √r = p/q As p is an integer, p² = p×p is also an integer, let y = p² And as q is an integer, q² = q×q is also an integer, let x = q² The number is the square of its square root, thus: (√r)² = (p/q)² = p²/q² = y/x but (√r)² = r, thus r = y/x and is a rational number. But r was chosen to be an irrational number, which is a contradiction (r cannot be both rational and irrational at the same time, so it cannot exist). Thus the square root of an irrational number cannot be rational. However, the square root of a rational number can be irrational, eg for the rational number ½ its square root (√½) is not rational.

Yes. The proof is easy. Let x be the irrational number and assume there exists some rational number r = a/b where a and b are integers (that's what it means to be rational). Now suppose x/r is a rational number. Then x/r = (b/a)x = c/d where c and d are some other integers. Since (b/a)x=c/d, then x = bd/ac which means that x itself is rational, but we assumed it was irrational. The contradiction proves that the assertion is wrong. An irrational divided by a rational must be irrational.

Yes! Every complex number z is a number, z = x + iy with x and y belonging to the field of real numbers. The real number x is called the real part and the real number y that accompanies i and called the imaginary part. The set of real numbers is formed by the meeting of the sets of rational numbers with all the irrational, thus taking only the complex numbers with zero imaginary part we have the set of real numbers, so then we have that for any irrational r is r real and complex number z = r + i0 = r and we r so complex number. So every irrational number is complex.

Minus pi. Or minus pi plus any rational number. Here is how you can figure this out (call your unknown number "x", and let "r" stand for any rational number):x + pi = r To solve for "x", simply subtract pi from both sides. That gives you: x = r - pi

There is no representation for irrational numbers: they are represented as real numbers that are not rational. The set of real numbers is R and set of rational numbers is Q so that the set of irrational numbers is the complement if Q in R.

Yes Yes, the sum of two irrational numbers can be rational. A simple example is adding sqrt{2} and -sqrt{2}, both of which are irrational and sum to give the rational number 0. In fact, any rational number can be written as the sum of two irrational numbers in an infinite number of ways. Another example would be the sum of the following irrational quantities [2 + sqrt(2)] and [2 - sqrt(2)]. Both quantities are positive and irrational and yield a rational sum. (Four in this case.) The statement that there are an infinite number of ways of writing any rational number as the sum of two irrational numbers is true. The reason is as follows: If two numbers sum to a rational number then either both numbers are rational or both numbers are irrational. (The proof of this by contradiction is trivial.) Thus, given a rational number, r, then for ANY irrational number, i, the irrational pair (i, r-i) sum to r. So, the statement can actually be strengthened to say that there are an infinite number of ways of writing a rational number as the sum of two irrational numbers.

The set of irrational numbers is NOT denoted by Q.Q denotes the set of rational numbers. The set of irrational numbers is not denoted by any particular letter but by R - Q where R is the set of real numbers.

There is no specific sign. The set of irrationals can be written as R - Q.