It is obviously true, right?

Let's take the irrational numbers $a+b\sqrt{c}$ and its conjugate $a-b\sqrt{c}$ , where $c$ is squarefree and $a,b$ are rational. If we add them, then we get the answer $2a$ which is rational.

Pi +(-Pi) =0 & 0is rational

We have to prove that irrational+irrational $\ne$ irrational.We can just put up an example to verify.E.g.. $(4-3\sqrt{3})+3\sqrt{3}=4$ which is a rational number.

consider the numbers √2 and -√2 both are irrationals but sum is 0 is rational

1/2(irrational)+1/2(irrational)=2/2=1 (rational)

root5 + -root5 = 0: 7 + root 3 = 8 - root3 = 15 ia rational

This is way more complicated than it needs to be, but imagine the numbers 0.45445444544445... and 0.54554555455554...

Their sum would be 1, which is rational.

Seriously? How about pi plus -pi... It equals zero... This wasn't a difficult question.

Considering that x is an irrational number between 0 and 1 and that a sum of an irrational number with a rational number is a rational number: x + (1-x) = 1, which is a rational number. Therefore, a sum of irrational numbers is not always an irrational number.

whenever we probe something is not true we need only one counter example and the example is i+(-i)=0 which is a rational number

If the sum of 2 irrational numbers is always an irrational number, then $\sqrt{2}+(2-\sqrt{2})$ must be an irrational number, but it is $2$ , which is rational, so it is false that the sum of 2 irrational numbers is always an irrational number.

You can add together an irrational and its additive inverse to make $0$ which is rational.

Proof: $a+(-a)=a-a=0$

The title kind of gave it away actually.

Pi + (1-Pi) = 1. Pi is irrational (and transcendental btw). (1-Pi) is irrational (again transcendental btw). 1 is not irrational (nor trancendental).

It's not always true. But for any irrational number, we can choose an irrational number such that their sum could be a rational number.