It is a rational number.
The sum or the difference between two irrational numbers could either be rational or irrational, however, it should be a real number.
A rational number is one that can be expressed as a/b The sum of two rational numbers is: a/b + c/d =ad/bd + bc/bd =(ad+bc)/bd =e/f which is rational The difference of two rational numbers is: a/b - c/d =ab/bd - bc/bd =(ab-bc)/bd =e/f which is rational The product of two rational numbers is: (a/b)(c/d) =ac/bd =e/f which is rational
There are an infinite number of rational numbers between any two rational numbers.
No, and I can prove it: -- The product of two rational numbers is always a rational number. -- If the two numbers happen to be the same number, then it's the square root of their product. -- Remember ... the product of two rational numbers is always a rational number. -- So the square of a rational number is always a rational number. -- So the square root of an irrational number can't be a rational number (because its square would be rational etc.).
Yes.
Yes. This is the same as asking for one rational number to be subtracted from another; to do this each rational number is made into an equivalent rational number so that the two rational numbers have the same denominator, and then the numerators are subtracted which gives a rational number which may possibly be simplified.
Yes, that's true.
Yes. The rational numbers are a closed set with respect to subtraction.
There is no number which can be rational and irrational so there is no point in asking "how".
The question cannot be answered because it is nonsensical. The difference between two rational numbers is very very rarely a whole number.
No. 5 and 2 are real numbers. Their difference, 3, is a rational number.
The difference of two rational numbers is rational. Let the two rational numbers be a/b and c/d, where a, b, c, and d are integers. Any rational number can be represented this way. Their difference is a/b-c/d = ad/bd-cb/bd = (ad-cb)/bd. Products and differences of integers are always integers. This means that ad-cb is an integer, and so is bd. Thus, (ad-cb)/bd is a rational number (since it is the ratio of two integers). This is equivalent to the difference of the original two rational numbers.
no
Yes.
Find the arithmetic average of the two rational numbers. It will be a rational number and will be between the two numbers.
Every time. The sum of two rational numbers MUST be a rational number.