Find $19\Diamond 98$, given rules $(xy)\Diamond y=x(y\Diamond y)$, $(x\Diamond 1)\Diamond x=x\Diamond 1$, and $1\Diamond 1=1$

Question

Given any two positive real numbers $x$ and $y$, then $x \, \Diamond \, y$ is a positive real number defined in terms of $x$ and $y$ by some fixed rule. Suppose the operation $x \, \Diamond \, y$ satisfies the equations $(xy) \, \Diamond \, y=x(y \, \Diamond \, y)$ and $(x \, \Diamond \, 1) \, \Diamond \, x = x \, \Diamond \, 1$ for all $x,y>0$.

Given that $1 \, \Diamond \, 1=1$, find $19 \, \Diamond \, 98$.

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I have tried solving this where $x\diamond y = f(x,y)$, but I couldn't get anywhere with it. Can anyone help?

Thanks!

Answer 1

Hint: you should be able to show

• $x \, \Diamond \, 1=x$ and
• $x \, \Diamond \, x=x$ and
• $(xy) \, \Diamond \, y = xy$ and
• $x \, \Diamond \, y = x$ (using $y \not=0$)

and thus

• $19 \, \Diamond \, 98 = 19$