show that √2 + √3 is an irrational
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Answer:
Let us suppose that √2+√3 is rational.
Let √2+√3=\frac{a}{b}
b
a
,
where a,b are integers and b≠0
Therefore,
\sqrt{2}=\frac{a}{b}-\sqrt{3}
2
=
b
a
−
3
On Squaring both sides , we get
2=\frac{a^{2}}{b^{2}}+3-2\times\frac{a}{b}\times\sqrt{3}2=
b
2
a
2
+3−2×
b
a
×
3
Rearranging the terms ,
\frac{2a}{b}\times\sqrt{3}=\frac{a^{2}}{b^{2}}+3-2
b
2a
×
3
=
b
2
a
2
+3−2
= \frac{a^{2}}{b^{2}}+1=
b
2
a
2
+1
\sqrt{3}=\frac{a^{2}+b^{2}}{2ab}
3
=
2ab
a
2
+b
2
Since , a,b are integers ,
\frac{a^{2}+b^{2}}{2ab}
2ab
a
2
+b
2
is rational, and so √3 is rational.
This contradicts the fact √3 is irrational.
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