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How you prove that -if r is rational and x is irrational then prove that r x and rx are rational?
To prove that if (r) is rational and (x) is irrational, then both (rx) and (\frac{r}{x}) are rational, we can use the fact that the product or quotient of a rational and an irrational number is always irrational. Since (r) is rational and (x) is irrational, their product (rx) must be irrational. Similarly, the quotient (\frac{r}{x}) must also be irrational. Therefore, we cannot prove that both (rx) and (\frac{r}{x}) are rational based on the given information.
What else can I help you with?
Product of irrational and rational?
The product is still irrational. Proof by contradiction: Let X and Y be an element of Q (the set of rational numbers). Thus xy can be written in the form p/q, where p, q are an element of Z (the set of integers), and q is NOT zero. Since X is rational (given in question), it can be written as r/s. Again, r,s are elements of Z and s is not equal to zero. This gives Y=ps/qr, which is also an element of Q (rationals). However, we know that Y is not rational (given in the question), hence this is the contradiction. Hence the product of a rational and an irrational is irrational.
Is every irrational number a complex number?
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.
Prove that if a and b are rational numbers then a multiplied by b is a rational number?
If a is rational then there exist integers p and q such that a = p/q where q>0. Similarly, b = r/s for some integers r and s (s>0) Then a*b = p/q * r/s = (p*r)/(q*s) Now, since p, q r and s are integers, p*r and q*s are integers. Also, q and s > 0 means that q*s > 0 Thus a*b can be expressed as x/y where p and r are integers implies that x = p*r is an integer q and s are positive integers implies that y = q*s is a positive integer. That is, a*b is rational.
What is the definition of real positive numbers?
A positive real number is any natural, integer, rational, or irrational number x such that x>0. In other words, the real numbers indicated by with or without positive sign (+) is known as Positive Real Number. Positive Real numbers are indicated by R+ mathematically.For example R+ = {1, 2, 3, 4, 5, .....}
Is 0.636363 an rational number?
yes... r= 0.63636363... 100r= 63.636363... 100r - r= 99r 99r = 63 r= 63/99 hope this helps!
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.
Why the area of a circle with a rational radius must be an irrational number?
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.
How do you write an irrational number in algebra?
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.
Is an irrational number divided by a rational number always irrational?
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.
Is the square root of an irrational number 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.
Can you add two irrational numbers to get a rational number?
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.
What irrational number can be added to pi to get a sum that is rational?
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
What are polynomials that can be factored?
A polynomial can be factored if it has a rational root. If f(x) is a polynomial function of x and if there is a rational number p such that f(p) = 0 then f(x) = (x-p)*g(x) where g(x) is a polynomial whose order is one less than the order of f(x). If p = q/r where q and r are integers, then (x - p) = (x - q/r) = (rx - q)/r which is a rational binomial factor. This does not work if p is irrational which is why p must be rational.
Why the set of irrational number is denoted by q'?
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.
Set of numbers made up of rational numbers and irrational numbers?
The real set, denoted R or ℝ.
Product of irrational and rational?
The product is still irrational. Proof by contradiction: Let X and Y be an element of Q (the set of rational numbers). Thus xy can be written in the form p/q, where p, q are an element of Z (the set of integers), and q is NOT zero. Since X is rational (given in question), it can be written as r/s. Again, r,s are elements of Z and s is not equal to zero. This gives Y=ps/qr, which is also an element of Q (rationals). However, we know that Y is not rational (given in the question), hence this is the contradiction. Hence the product of a rational and an irrational is irrational.
Is 0.757575 rational or irrational?
The number 0.757575 is a rational number. A rational number is any number that can be expressed as a fraction where the numerator and denominator are integers and the denominator is not zero. In this case, 0.757575 can be expressed as the fraction 75/99, which meets the criteria for a rational number.
