There can be no such number.
2 times pi is not an integer. Since Pi is an irrational number, 2 pi is also an irrational number.
No, a real number could also be a rational number, an integer, a whole number, or a natural number. Irrational numbers fall into the same category of real numbers, but every real number is not an irrational number.
Absolutely. Only fractions can be irrational, numerically speaking (people can also be irrational, but that's a different use of the word).
No. If a number is irrational, it continues endlessly without a pattern. Since 2.5 stops at 5, it is rational; but if it were, say, 2.573583..., it would be irrational. Also, it can be written as the fraction 25/10, or 5/2.
There can be no such number.
2 times pi is not an integer. Since Pi is an irrational number, 2 pi is also an irrational number.
No, it is rational; it is also an integer.
Since pi is an irrational number, then 3pi will be an irrational also.
21 is an integer which is also a rational number because it can be expressed as a fraction in the form of 21/1
No, a real number could also be a rational number, an integer, a whole number, or a natural number. Irrational numbers fall into the same category of real numbers, but every real number is not an irrational number.
Actually there are more irrational numbers than rational numbers. Most square roots, cubic roots, etc. are irrational (not rational). For example, the square of any positive integer is either an integer or an irrational number. The numbers e and pi are both irrational. Most expressions that involve those numbers are also irrational.
No, it is a real, rational number. It can also be called an integer, whole number, and natural number.
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Absolutely. Only fractions can be irrational, numerically speaking (people can also be irrational, but that's a different use of the word).
No. If a number is irrational, it continues endlessly without a pattern. Since 2.5 stops at 5, it is rational; but if it were, say, 2.573583..., it would be irrational. Also, it can be written as the fraction 25/10, or 5/2.
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.