Answer:
√3+√7 is irrational.
Step-by-step explanation:
Let us assume that √3+√7 is rational.
That is , we can find coprimes a and b (b≠0) such that \sqrt{3}+\sqrt{7}=\sqrt{a}{b}
3
+
7
=
a
b
Therefore,
\sqrt{7}=\frac{a}{b}-\sqrt{3}
7
=
b
a
−
3
Squaring on both sides ,we get
7=\frac{a^{2}}{b^{2}}+3-2\times \frac{a}{b}\times \sqrt{3}7=
b
2
a
2
+3−2×
b
a
×
3
Rearranging the terms ,
\begin{gathered}2\times \frac{a}{b}\times \sqrt{3}=\frac{a^{2}}{b^{2}}+3-7\\=\frac{a^{2}}{b^{2}}-4\end{gathered}
2×
b
a
×
3
=
b
2
a
2
+3−7
=
b
2
a
2
−4
\implies 2\times \frac{a}{b}\times \sqrt{3}=\frac{a^{2}-4b^{2}}{b^{2}}⟹2×
b
a
×
3
=
b
2
a
2
−4b
2
\begin{gathered}\implies \sqrt{3}=\frac{a^{2}-4b^{2}}{b^{2}}\times \frac{b}{2a}\\=\frac{a^{2}-4b^{2}}{2ab}\end{gathered}
⟹
3
=
b
2
a
2
−4b
2
×
2a
b
=
2ab
a
2
−4b
2
Since, a and b are integers , \frac{(a^{2}-4b^{2})}{2ab}
2ab
(a
2
−4b
2
)
is rational ,and so √3 also rational.
But this contradicts the fact that √3 is irrational.
This contradiction has arisen because of our incorrect assumption that √3+√7 is rational.
Hence, √3+√7 is irrational.
