Friends kindly enlighten me on the following aspects:
Is 3 + Sq. root of 2 - irrational?
Whether irrational roots occur in pair?
As an example (3+ Sq. root of 2) and (3 - Sq. root of 2)
Hope I have made myself clear.
Thanks.
$\displaystyle 3+\sqrt{2}$ is an irrational number.
If $\displaystyle \mathbf{r}$ is rational and $\displaystyle \gamma$ is irrational then $\displaystyle \mathbf{r}+\gamma$ is irrational.
If the coefficients of a quadratic are rational then any irrational root has a pairing.
If all coefficents of a quadratic equation are rational, then the all irrational roots appear in "conjugate pairs". Your statement, "So the coefficients of a quadratic should be rational for the roots to be (irrational) pair", however, is not necessarily true. An obvious example is $\displaystyle \sqrt{3}x^2- 5\sqrt{3}x+ 2\sqrt{3}= 0$ which has irrational coefficients but "irrational pair" roots, $\displaystyle \frac{1\pm\sqrt{17}}{2}$.
Suppose j is any rational number, in this case 3. I am a math beginner, but since j + any rational number is equal to a rational number [j + rational number = rational number], 3 + any rational number = a rational number. I know obvious. Now comes the challenge, is 3 + an irrational number equal to an irrational number? When we found out that 3 + any rational number = a rational number, so 3 + any irrational number can't be a rational number or j + irrational number does not equal a rational number. Therefore, j + an irrational number must be irrational also.