√19 + √3 is greater than √5 +√13.
We first relate the two numbers √19 + √3 and √5 +√13 through an inequality and check if it is true.
First suppose that √19 + √3 > √5 +√13
On squaring both sides and using the identity [tex](a+b)^2=a^2+b^2+2ab[/tex],
⇒ [tex](\sqrt{19}+\sqrt{3})^2 > (\sqrt{5}+\sqrt{13})^2\\22+2\sqrt{57} > 18+2\sqrt{65} \\[/tex]
Dividing by 2 on both sides,
⇒ [tex]11+\sqrt{57} > 9+\sqrt{65}[/tex]
Subtracting 9 from both sides,
⇒ [tex]2+\sqrt{57} > \sqrt{65}[/tex]
Squaring again, we get,
⇒ [tex]61 +4\sqrt{57} > 65[/tex]
Subtract 61 on both sides,
⇒ [tex]4\sqrt{57} > 4[/tex]
Dividing by 4 on both sides,
⇒ [tex]\sqrt{57} > 1[/tex]
Finally squaring once more,
⇒ 57 > 1, which is true.
So our assumption that √19 + √3 > √5 +√13 is right.
Learn more about irrationals at https://brainly.com/question/20400557
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