This can easily be proved by contradiction. Without loss of generality, I will take specific numbers as an example. The proof can easily be extended to any rational + irrational number.
Assumption: 1 plus the square root of 2 is rational. (It is a well-known fact that the square root of 2 is irrational. No need to prove it here; you can use any other irrational number will do.)
This rational sum can be written as p / q, where "p" and "q" are whole numbers (this is basically the definition of a "rational number").
Then, the square root of 2, which is equal to the sum minus 1, is:
p / q - 1
= p / q - q / q
= (p - q) / q
Since the difference of two whole numbers is a whole number, this makes the square root of 2 rational, which doesn't make sense.
No. In fact the sum of a rational and an irrational MUST be irrational.
The value of the sum depends on the values of the rational number and the irrational number.
It is always irrational.
The sum is irrational.
An irrational number.
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.
The value of the sum depends on the values of the rational number and the irrational number.
It is always irrational.
The sum is irrational.
Such a sum is always irrational.
An irrational number.
The sum of the three can be rational or irrational.
Any, and every, irrational number will do.
Since the sum of two rational numbers is rational, the answer will be the same as for the sum of an irrational and a single rational number. It is always irrational.
The sum of two irrational numbers may be rational, or irrational.
Yes