Question

Let S be the set of all points $\left(\frac{a}{b},\frac{c}{d}\right)$ on the circle with radius $1$ centred at $(0,0)$ where a and b are relatively prime integers, c and d are relatively prime integers (that is HCF(a, b)$=$HCF(c, d)$=1$), and the integers b and d are even. Then the set S.

• A. Is empty
• B. Has four elements
• C. Has eight elements
• D. Is infinite

Answer 1

The correct option is A Is empty

Given that $b,d$ are even,

So let us assume that $b=2^{p1}k,d=2^{p2}l(p1,p2>=1,k,l$ are odd$)$

Also we know that $a,b$ are coprimes$⟹$a is odd $(\because$ b is even$)$

Similarly , $c$ is odd

Now given that,$(\frac{a}{b},\frac{c}{d})$ satisfies the equation of a circle centre at origin and radius 1

That is $x^2+y^2=1⟹\frac{a^2}{b^2}+\frac{c^2}{d^2}=1⟹\frac{a^2}{2^{2p1}{k}^2}+\frac{c^2}{2^{2p2}{l}^2}=1⟹2^{2p2}{a}^2{l}^2+2^{2p1}{c}^2{k}^2=2^{2p1+2p2}{k}^2{l}^2⟹a^2{l}^2+2^{2p1-2p2}{c}^2{k}^2=2^{2p1}{k}^2{l}^2$

Now if $p1=p2$then

$a^2{l}^2+c^2{k}^2=2^{2p1}{k}^2{l}^2$

But $a,k,l,c$ are odd $⟹a^2{l}^2,c^2{k}^2$ are odd $⟹a^2{l}^2+b^2{k}^2$ is a multiple of two

But in RHS ,$p1>=1⟹2p1>=2⟹2^{2p1}>=4$

$⟹$ RHS is atleast a multiple of 4 but not 2

Hence,$p1\ne p2$

If $p1>p2$,

Then $a^2{l}^2$ is odd $2^{2p1-2p2}{c}^2{k}^2$ is even

So $a^2{l}^2+2^{2p1-2p2}{c}^2{k}^2$ is odd,but RHS is even

Hence, $p1\ngtr p2$

Similarly, we can prove that $p1\nless p2$ by dividing by $2^{2p1}$

Hence the set is empty.