3) Show that 5+√
7 is an irrational number
Answers
Answer:
Let us assume to the contrary that 3√5-7 is a rational no.
Such That ,
3√5-7=p/q {where, p and q are integers having no common factors}.
3√5=p/q+7
3√5=.p+7q/q
√5= p+7q/3q
where , √5 and p+7q/3q are rational numbers.
But this contradicts the fact that √5 is irrational.
Therefore,3√5-7 is an irrational no.
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let us assume that 5+√7 is an rational number
such that 5+√7=a/b (b≠0) . a and b are coprime integers
so,
5+√7=a/b
on squaring both side
(5+√7)² =a²/b²
(5)²+(√7)²+2×5×√7=a²/b²
25+7+10√7=a²/b²
32+10√7=a²/b²
10√7=a²/b²-32
10√7=(a²-32b²)/b²
√7=(a²-32b²)/7b²
hence, a and b are coprime integers
so, (a²-32b²)/7b² is rational number
then √7 is rational number
now, our assumption is wrong
so , 5+√7 is an irrational number