Step 1

Let \(\displaystyle{X}={\left\lbrace{1},\ {2},\ {3},\cdots,{60}\right\rbrace}\) with \(\displaystyle{\left|{X}\right|}={60}\) and

Let \(\displaystyle\phi={\left\lbrace{1},\ {7},\ {3},\cdots,{40}\right\rbrace}\) with \(\displaystyle{\left|\phi\right|}={31}\)

Claim: \(\displaystyle\exists\ {a},\ {b}\in\ {X}\) such that \(\displaystyle{\frac{{{b}}}{{{a}}}}={k}\in\ {X}\)

Suppose not then \(\displaystyle{a},\ {b}\in\ {X}\)

but if \(\displaystyle{b}={3}\) and \(\displaystyle{a}={2}\)

with \(\displaystyle{a}{<}{b}\) is not satisfying the condition \(\displaystyle{\frac{{{b}}}{{{a}}}}={k}\in\ {X}\)

also when \(\displaystyle{a}{>}{b},\) with \(\displaystyle{a}={3}\) and \(\displaystyle{b}={2}\) still it

is not satisflying the condition \(\displaystyle{\frac{{{b}}}{{{a}}}}={k}\in\ {X}\)

it satisflying the condition \(\displaystyle{\frac{{{b}}}{{{a}}}}={k}\in\ {X}\)

only if \(\displaystyle{a}={b}\)

therefore, \(\displaystyle\exists{a},\ {b}\in{X}\) such that \(\displaystyle{\frac{{{b}}}{{{a}}}}={k}\in\ {X}\), only if \(\displaystyle{a}={b}\)