Math, asked by vishal2004jayapak1jf, 10 months ago

Prove that 3√5-7 is an irrational number. ​

Answers

Answered by vikeshjee
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 Anonymous
0

3√5-7=a/b

3√5=a/b+7

√5 = a+7b/3b

Hence, proved...

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