an irrational number PROOF : Let x be any rational number and y be any irrational number. let us assume that their sum is rational which is ( z ) x + y = z if x is a rational number then ( -x ) will also be a rational number. Therefore, x + y + (-x) = a rational number this implies that y is also rational BUT HERE IS THE CONTRADICTION as we assumed y an irrational number. Hence, our assumption is wrong. This states that x + y is not rational. HENCE PROVED
it will always be irrational.
Let R1 = rational number Let X = irrational number Assume R1 + X = (some rational number) We add -R1 to both sides, and we get: -R1 + x = (some irrational number) + (-R1), thus X = (SIR) + (-R1), which implies that X, an irrational number, is the sum of two rational numbers, which is a contradiction. Thus, the sum of a rational number and an irrational number is always irrational. (Proof by contradiction)
Yes
No. The sum of an irrational number and any other [real] number is irrational.
That simply isn't true. The sum of two irrational numbers CAN BE rational, but it can also be irrational. As an example, the square root of 2 plus the square root of 2 is irrational.
Suppose x is a rational number and y is an irrational number.Let x + y = z, and assume that z is a rational number.The set of rational number is a group.This implies that since x is rational, -x is rational [invertibility].Then, since z and -x are rational, z - x must be rational [closure].But z - x = y which implies that y is rational.That contradicts the fact that y is an irrational number. The contradiction implies that the assumption [that z is rational] is incorrect.Thus, the sum of a rational number x and an irrational number y cannot be rational.
Yes. The sum of two irrational numbers can be rational, or irrational.
The sum of a rational and irrational number must be an irrational number.
No. In fact the sum of a rational and an irrational MUST be irrational.
Not necessarily. The sum of two irrational numbers can be rational or irrational.
The value of the sum depends on the values of the rational number and the irrational number.
Let `a` be a rational number and `b` be an irrational number,assume that the sum is rational. 1.a +b =c Where a and c are rational and b is irrational. 2.b=c-a Subtracting the same number a from each side. 3.b is irrational c-a is a rational number we arrived at a contradiction. So the sum is an irrational number.
Let R1 = rational number Let X = irrational number Assume R1 + X = (some rational number) We add -R1 to both sides, and we get: -R1 + x = (some irrational number) + (-R1), thus X = (SIR) + (-R1), which implies that X, an irrational number, is the sum of two rational numbers, which is a contradiction. Thus, the sum of a rational number and an irrational number is always irrational. (Proof by contradiction)
It is always irrational.
The sum is irrational.
Such a sum is always irrational.
An irrational number.
Yes Yes, the sum of two irrational numbers can be rational. A simple example is adding sqrt{2} and -sqrt{2}, both of which are irrational and sum to give the rational number 0. In fact, any rational number can be written as the sum of two irrational numbers in an infinite number of ways. Another example would be the sum of the following irrational quantities [2 + sqrt(2)] and [2 - sqrt(2)]. Both quantities are positive and irrational and yield a rational sum. (Four in this case.) The statement that there are an infinite number of ways of writing any rational number as the sum of two irrational numbers is true. The reason is as follows: If two numbers sum to a rational number then either both numbers are rational or both numbers are irrational. (The proof of this by contradiction is trivial.) Thus, given a rational number, r, then for ANY irrational number, i, the irrational pair (i, r-i) sum to r. So, the statement can actually be strengthened to say that there are an infinite number of ways of writing a rational number as the sum of two irrational numbers.