Question

prove √7 is ir-rational number​

Answer 1

$\large{\underline{\underline{\red{\sf{Given:}}}}}$

• $\tt{A\: irrational\:number\:is\:given.}$

$\large{\underline{\underline{\red{\sf{To\:Prove:}}}}}$

• $\tt{\sqrt{7}\:is\:a\: irrational\:number.}$

$\large{\underline{\underline{\red{\sf{Proof:}}}}}$

$\tt{Given\:number\:is\:\sqrt{7}.}$

So , on the contrary let us assume that √7 is a Rational number . So it can be expressed in the form of p/q , where p and q are integers and q ≠ 0. Also HCF of p & q is 1 which means p and q are co - primes.

$\underline{\purple{\rm{\leadsto As\:per\:our\: assumption:- }}}$

$\tt{\implies \sqrt{7}=\dfrac{p}{q}}$

$\tt{\implies (\sqrt{7})^2=(\dfrac{p}{q})^2}$

$\tt{\implies 7 = \dfrac{p^2}{q^2}}$

$\green{\tt{\implies p^2=7q^2}}$

• This implies that 7 is a factor of p² .So by "Fundamental Theorem of Arthemetic " we can say that 7 is a factor of p also.

$\orange{\tt{Let\: p = 7k}}$

$\pink{\sf{Putting \:this\: value\:in\:above\:equ^n}}$

$\tt{\implies p^2=7q^2 }$

$\tt{\implies (7k)^2=7q^2}$

$\tt{\implies 49k^2=7q^2}$

$\tt{\implies q^2=\dfrac{49k^2}{7}}$

$\tt{\implies q^2 = 7k^2 }$

• This implies that 7 is a factor of q² .So by "Fundamental Theorem of Arthemetic " we can say that 7 is a factor of q also.

But this contradicts our assumption that p and q are co- primes , since we found that 7 is also a factor of p and q .

Hence our assumption was wrong .√7 must not be a Rational number .

$\red{\rm{\underset{\green{Hence\:Proved}}{\underbrace{\mapsto Hence \:\sqrt{7}\:is\:a\: irrational\:number.}}}}$

_________________________________

$\large{\underline{\underline{\red{\tt{\dag Extra\: Information:}}}}}$

$\underline{\purple{\rm{\leadsto Some\:properties\:of\: Irrational\:number :- }}}$

$\orange{\tt{\mapsto Sum\:of\: Rational\:and\: Irrational\:is\:Irrational. }}$

$\orange{\tt{\mapsto Sum\:of\:two\: Irrational\:may\:or\:may\:not\:be\:Irrational. }}$

$\orange{\tt{\mapsto Product \:of\: Rational\:and\: Irrational is Irrational}}$

$\orange{\tt{\mapsto Product\:of\:two\: Irrational\:may\:or\:may\:not\:be\: Irrational }}$

$\orange{\tt{\mapsto Difference\:of\: Rational\:and\: Irrational\:is\:Irrational. }}$

$\orange{\tt{\mapsto Difference\:of\:two\: Irrational\:may\:or\:may\:not\:be\:Irrational. }}$

$\orange{\tt{\mapsto Division\:of\: Rational\:and\: Irrational\:is\:Irrational. }}$

$\orange{\tt{\mapsto Division\:of\:two\: Irrational\:may\:or\:may\:not\:be\:Irrational. }}$