If `{x}` and `[x]` represent fractiona and integral part of `x` respectively then solve the equation `x-1=(x-[x])(x-{x})`

Explanation

`:'x=[x]+{x},0le{x}le1`
Thus given equation reduces to
`[x]+{x}=1={x}[x]`
`implies{x}[x]-[x]-{x}+1=0`
`implies([x]-1)({x}-1)=0`
Now `{x}-1!=0 [ :' 0 le {x}lt1]`
`:.[x]-1=0`
`implies[x]=1`
`:.x epsilon [1,2)`