The conjugate of an irrational number is a non-zero number such that the product of the two numbers is rational.
In high school mathematics, you are usually required to know only the conjugates of surds of the form a + b*sqrt(x).
The conjugate is a - b*sqrt(x)
[-a + b*sqrt(x) is also a conjugate - all you need to do is to switch the sign of one of the two terms.]
If the square root of a number is irrational, it is its own conjugate. sqrt(13)*sqrt(13) = 13 and you no longer have an irrational!
It depends on what the denominator was to start with: a surd or irrational or a complex number. You need to find the conjugate and multiply the numerator by this conjugate as well as the denominator by the conjugate. Since multiplication is by [conjugate over conjugate], which equals 1, the value is not affected. If a and b are rational numbers, then conjugate of sqrt(b) = sqrt(b) conjugate of a + sqrt(b) = a - sqrt(b), and conjugate of a + ib = a - ib where i is the imaginary square root of -1.
No. In fact, the sum of conjugate irrational numbers is always rational.For example, 2 + sqrt(3) and 2 - sqrt(3) are both irrational, but their sum is 4, which is rational.
No. The sum of an irrational number and any other [real] number is irrational.
A negative irrational number can be thought of as an irrational number multiplied by -1, or an irrational number with a minus sign in front of it.
If the square root of a number is irrational, it is its own conjugate. sqrt(13)*sqrt(13) = 13 and you no longer have an irrational!
It depends on what the denominator was to start with: a surd or irrational or a complex number. You need to find the conjugate and multiply the numerator by this conjugate as well as the denominator by the conjugate. Since multiplication is by [conjugate over conjugate], which equals 1, the value is not affected. If a and b are rational numbers, then conjugate of sqrt(b) = sqrt(b) conjugate of a + sqrt(b) = a - sqrt(b), and conjugate of a + ib = a - ib where i is the imaginary square root of -1.
No. The product of sqrt(2) and sqrt(2) is 2, a rational number. Consider surds of the form a+sqrt(b) where a and b are rational but sqrt(b) is irrational. The surd has a conjugate pair which is a - sqrt(b). Both these are irrational, but their product is a2 - b, which is rational.
No. In fact, the sum of conjugate irrational numbers is always rational.For example, 2 + sqrt(3) and 2 - sqrt(3) are both irrational, but their sum is 4, which is rational.
Graphically, the conjugate of a complex number is its reflection on the real axis.
An irrational number.
No. The sum of an irrational number and any other [real] number is irrational.
The sum of a rational and irrational number must be an irrational number.
No, 3.56 is not an irrational number. 3.56 is rational.
rational * irrational = irrational.
-Pi is irrational, because it does not terminate or repeat. Whenever you multiply an irrational number by a rational number (-1), the result is an irrational number.
When a complex number is multiplied by its conjugate, the product is a real number and the imaginary number disappears.