Question

prove that 7+3 root 5 is irrational number​

Answer 1

Step-by-step explanation:

Letus assume the contrary 7+3root 5 is rational no.

So we find coprime integer a and b where b is not equal to 0

7+3√5= a/b

3√5=a/b-7

√5=a-7b/3b

√5is irrational no.

But this contradicts the fact that √5is irrational

Our assumptions is wrong

Therefore 7+3√5is irrational

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Answer 2

$\huge{\mathcal{\blue{\underline{\underline{\pink{HeYa!!!}}}}}}$

${\large{\purple{\mathbb{ANSWER}}}}$

${\large{\blue{\bold{\underline{To\:Prove:-}}}}}$

▪︎ 7+3root5 is irrational.

${\large{\blue{\bold{\underline{Proof,}}}}}$

▪︎ Let us assume it as a rational number.

${\large{\blue{\bold{\underline{Therefore,}}}}}$

=》 7+3 root5 = p/q,

Where p and q are integers and q is not equal to 0 alsp HCF(p,q) = 1.

=》 3 root5 = p/q - 7.

=》 3 root5 = p-7q/q.

{By taking LCM}

=》 root 5 = p-7q/3q

▪︎ Also,

=》 p-7q/3q = Integer/Integer.

Therefore it is a rational number.

=》 root 5 is rational.

${\large{\blue{\bold{\underline{But,}}}}}$

▪︎ It contradicts the fact that root 5 is irrational.

▪︎ This contradiction has arisen because of our wrong assumption.

▪︎ So our assumption that 7+3root5 is a rational number was wrong.

▪︎ Therefore we can conclude that 7+3root5 is an irrational number.

Hence proved.

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