Mathematics I - Old Questions

An irrational number is a number that cannot be expressed as a fraction  for any integers  and .

Let us assume on the contrary that $\sqrt{2}$ is a rational number. Then, there exist positive integers p and q such that

 , where, p and q, are co-prime i.e. their $HCF$ is $1.$

$On\; squaring\; both\; sides,\; we\; get$

$p²=\; 2q²\; ......\; (1)$

$Clearly,\; 2\; is\; a\; factor\; of\; 2q²\Rightarrow \; 2\; is\; a\; factor\; of\; p²\; [since,2q²=p²]\Rightarrow \; 2\; is\; a\; factor\; of\; p$

$Let\; p\; =2m\; for\; all\; m\; (\; where\; m\; is\; a\; positive\; integer)Squaring\; both\; sides,\; we\; getp²=\; 4\; m²\; .....(2)$

$From\; (1)\; and\; (2),\; we\; get2q²\; =\; 4m²\; \Rightarrow \; q²=\; 2m²Clearly,\; 2\; is\; a\; factor\; of\; 2m²\Rightarrow 2\; is\; a\; factor\; of\; q²\; [since,\; q²\; =\; 2m²]\Rightarrow \; 2\; is\; a\; factor\; of\; q$

$Thus,\; we\; see\; that\; both\; p\; and\; q\; have\; common\; factor\; 2\; which\; is\; a\; contradiction\; that\; H.C.F.\; of\; (p,q)=\; 1Therefore,\; Our\; supposition\; is\; wrongHence\; \surd 2\; is\; not\; a\; rational\; number\; i.e.,\; irrational\; number.$