Proof : ax+bcx+d is one to one correspondence
$ \displaystyle f(x)=ax+bcx+df is one to one ⇒f(x)=f(y)⇒x=y Assume that f(x)=f(y),thenax+bcx+d=ay+bcy+dacxy+adx+bcy+bd=acxy+bcx+ady+bdadx+bcy=bcx+adyadx−bcx=ady−bcy(ad−bc)x=(ad−bc)y∴ x=y (ad≠bc)Hence f is one to one.$
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