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.
No. A rational plus an irrational is always an irrational.
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.
rational
It is always irrational.
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.
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.
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.
4.6 is rational.
10.01 is a rational number
Rational
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. 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 34.54 and irrational or rational. number
Rational
Rational.