Prove that 3√5-7 is an irrational number.
Answers
Answered by
1
Answer:
Step-by-step explanation:
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.
Answered by
0
3√5-7=a/b
3√5=a/b+7
√5 = a+7b/3b
Hence, proved...
Similar questions