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Q: Can a number be rational and irrational the same time?

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It can't be both at the same time. Irrational means "not 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.

' 9 ' can certainly be one ... or many ... of the digits in a written approximation of an irrational number. The number ' 9 ' itself is a perfectly rational number. If that doesn't answer your question, that's because I'm having a tough time understanding your question.

For example, by taking the square root of any positive integer, except a perfect square. Thus, the square root of 2, 3, 5, 6, 7, 8, 10, 11, etc. are all irrational. You can also make up a rule to write a number in decimal, that does not involve a regular repetition. Note that, for example, 5.4871313131313... (repeating "13" forever after that) is rational. However, if you write, for example, 0.1010010001000010000010000001... (adding one more zero each time) will give you an irrational number, since all rational numbers will repeat the same sequence of digits over and over eventually.

Any time your decimal is not repeating, it is rational.

Related questions

Integers and fractions that have integers in the numerator and denominator are rational. A number can't be rational and irrational at the same time - irrational means "not rational".

Numbers cannot be rational and irrational at the same time.

No number can be rational and irrational at the same time. 3.14 is the ratio of 314:100 and so is rational. HOWEVER, 3.14 is also a common approximation for pi, which is an irrational number. All irrational numbers have infinite, non-recurring decimals and so are often approximated by rationals.

It can't be both at the same time. Irrational means "not rational".

Yes. Any time you multiply a rational number by an irrational number, you get an irrational number - unless the rational number is zero.

Every time. No exceptions.

A whole number k can be written in the form k/1 where k and 1 are both integers. It can, thus, be expressed in the form of a ratio and so is rational. Since it is rational it cannot be irrational. Simple!

None. A rational number is a number that can be written as the quotient of two integers where the divisor is not zero. An irrational number is a real number that cannot be written as the quotient of two integers where the divisor is not zero. Any given real number either can or cannot be written as the quotient of two integers. If it can, it is rational. If it cannot, it is irrational. You can't be both at the same time. The square root of -1 is not a real number and it cannot be written as the quotient of two integers, so it is neither rational nor irrational.

Any time a pattern of digits repeats over and over, it's a rational number.

No they cant because that would be contradicting each other ( The numbers wont end and don't have a pattern but rational is the complete opposite)

12 is a rational number. Incidentally, there is no such time as 12 am or 12 pm: they are 12 noon and 12 midnight. Furthermore, if it was a time then it would not be a number (numbers do not have measurement units associated with them) and so it could not 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.

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