True.
It is true.
The question cannot be answered because it is based on a false premise.The product of a (not an!) rational number and an irrational number need not be irrational. For eample, the product ofthe rational number, 0, and the irrational number, pi, is 0. The product is rational, not irrational!
The sum of a rational and an irrational number is always irrational. Here is a brief proof:Let a be a rational number and b be an irrational number, and c = a + b their sum. By way of contradiction, suppose c is also rational. Then we can write b = c - a. But since c and a are both rational, so is their difference, and this means that bis rational as well. But we already said that b is an irrational number. This is a contradiction, and hence the original assumption was false. Namely, the sum c must be an irrational number.
Let x be a rational number and y be an irrational number.Suppose their sum = z, is rational.That is x + y = zThen y = z - xThe set of rational number is closed under addition (and subtraction). Therefore, z - x is rational.Thus you have left hand side (irrational) = right hand side (rational) which is a contradiction.Therefore, by reducio ad absurdum, the supposition that z is rational is false, ie the sum of a rational and an irrational must be irrational.
It is false.
It is true.
The question cannot be answered because it is based on a false premise.The product of a (not an!) rational number and an irrational number need not be irrational. For eample, the product ofthe rational number, 0, and the irrational number, pi, is 0. The product is rational, not irrational!
It is always FALSE.
In between any two rational numbers there is an irrational number. In between any two irrational numbers there is a rational number.
In between any two rational numbers there is an irrational number. In between any two Irrational Numbers there is a rational number.
The sum of a rational and an irrational number is always irrational. Here is a brief proof:Let a be a rational number and b be an irrational number, and c = a + b their sum. By way of contradiction, suppose c is also rational. Then we can write b = c - a. But since c and a are both rational, so is their difference, and this means that bis rational as well. But we already said that b is an irrational number. This is a contradiction, and hence the original assumption was false. Namely, the sum c must be an irrational number.
Let x be a rational number and y be an irrational number.Suppose their sum = z, is rational.That is x + y = zThen y = z - xThe set of rational number is closed under addition (and subtraction). Therefore, z - x is rational.Thus you have left hand side (irrational) = right hand side (rational) which is a contradiction.Therefore, by reducio ad absurdum, the supposition that z is rational is false, ie the sum of a rational and an irrational must be irrational.
It is false.
False.
Let q be a non-zero rational and x be an irrational number.Suppose q*x = p where p is rational. Then x = p/q. Then, since the set of rational numbers is closed under division (by non-zero numbers), p/q is rational. But that means that x is rational, which contradicts x being irrational. Therefore the supposition that q*x is rational must be false ie the product of a non-zero rational and an irrational cannot be rational.
False because 0.8 is a rational number which can be expressed as a fraction of 4/5 in its simplest form
It is true and false. It cannot be proved.