maths

If $a=c$, then $a+bβ=c+dββbβ=dββb=d$

So, let $aξ=c$. Then, there exists a positive rational number $x$ such that $a=c+x$.

Now,

$βa+bβ=c+dβ$

$βc+x+bβ=c+dβ$ $[β΅a=c+x]$

$βx+bβ=dβ$

$β(x+bβ)_{2}=(dβ)_{2}$

$βx_{2}+2bβx+b=d$

$βbβ=2xdβx_{2}βbβ$

$βbβ$ is rationalΒ Β $[β΄d,x,bΒareΒrationalsβ΄2xdβx_{2}βb_{2}βisΒrational]$

$β$ $b$ is the square of a rational number.

From$(i)$, we have

$dβ=x+bβ$Β

$β$ $dβ$ is rational

$β$ $d$ is the square of a rational number.

Hence, either $a=c$ and $b=d$ or $b$ and $d$ are the squares of rationals.

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