Yes.
For example, the sum of 2 + √3 and 2 - √3 is 4.
The quotient of a nonzero rational number and an irrational number is always an irrational number. This is because dividing a rational number (which can be expressed as a fraction of integers) by an irrational number cannot result in a fraction that can be simplified to a rational form. Therefore, the result remains outside the realm of rational numbers.
It is an irrational number.
The product will be irrational.
Yes, always.
yes * * * * * No. Rational and irrational numbers are two DISJOINT subsets of the real numbers. That is, no rational number is irrational and no irrational is rational.
Actually the product of a nonzero rational number and another rational number will always be rational.The product of a nonzero rational number and an IRrational number will always be irrational. (You have to include the "nonzero" caveat because zero times an irrational number is zero, which is rational)
It is always irrational.
Yes, as long as the two nonzero numbers are themselves rational. (Since a rational number is any number that can be expressed as the quotient of two rational numbers, or any number that can be written as a fraction using only rational numbers.) If one of the nonzero numbers is not rational, the quotient will most likely be irrational.
It is irrational.
An irrational number.
It is an irrational number.
Yes.
The product will be irrational.
Yes, always.
Yes, always.
The product of an irrational number and a rational number, both nonzero, is always irrational
No. If we let x be irrational, then 0/x = 0 is a counterexample. However, if we consider nonzero rational numbers, then our conjecture is true. We shall prove this by contradiction. Suppose we have nonzero rational numbers x and y, and an irrational number z, such that x/z = y. Since z is not equal to 0, x = yz. Since y is not equal to 0, x/y = z. Since x/y is a quotient of rational numbers, x/y is rational. Therefore, z is rational, contradicting our assumption that z was irrational. QED.