Given:
The sum of irrational number in the options.
To find:
The option that shows the sum of two irrational numbers can be irrational.
Solution:
In option A,
[tex](5+\pi)+(3-\pi)=5+\pi+3-\pi[/tex]
[tex](5+\pi)+(3-\pi)=5+3[/tex]
[tex](5+\pi)+(3-\pi)=8[/tex]
We know that 8 is a rational number. So, option A is incorrect.
In option B,
[tex]\sqrt{3}+\sqrt{5}[/tex]
Here both numbers are irrational and it cannot be simplified further.
So, [tex]\sqrt{3}+\sqrt{5}[/tex] is an irrational number and option B is correct.
In option C,
[tex](3+\sqrt{5})+(3-\sqrt{5})=3+\sqrt{5}+3-\sqrt{5}[/tex]
[tex](3+\sqrt{5})+(3-\sqrt{5})=3+3[/tex]
[tex](3+\sqrt{5})+(3-\sqrt{5})=6[/tex]
We know that 6 is a rational number. So, option C is incorrect.
In option D,
[tex](\dfrac{\pi}{3})+(-\dfrac{\pi}{3})=\dfrac{\pi}{3}-\dfrac{\pi}{3}[/tex]
[tex](\dfrac{\pi}{3})+(-\dfrac{\pi}{3})=0[/tex]
We know that 0 is a rational number. So, option D is incorrect.
Therefore, the correct option is B.
