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.
rational
It is always irrational.
The product of a rational and irrational number can be rational if the rational is 0. Otherwise it is always irrational.
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.
Rational
10.01 is a rational number
is 34.54 and irrational or rational. number
Rational
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.
No
If x is rational the it is rational. If x is irrational then it is irrational.