Find all values of x in the interval 1 <= x < 2 that satisfy [x]^2 = [x^2]. Note: [x] stands for the floor of x.
+121 let \(x=1 + r\), where r is the decimal part of x , \(0 . Then \(\lfloor x \rfloor=1\), and, therefore, \(\lfloor x \rfloor ^2=1\) . So we have the left side of the equation and we must look for conditions on r that give you \(\lfloor x^2 \rfloor=1\). But
\(x^2=(1+r)^2=1+2r+r^2\), so that for \(\lfloor x^2 \rfloor=\lfloor 1+2r+r^2\rfloor\) to equal 1, we must have the remainder of the expression, namely, \(2r+r^2\), be a number between 0 and 1, that is, \(0<2r+r^2<1\). We only have to consider, however, the less-than part of the inequality since r is a positive number.
For \(2r+r^2<1\) to hold, however, we can solve \(r^2+2r-1<0\) using a sign chart, but a better approach would be to notice that we are dealing with a parabola that opens up and that has vertex below the x-axis (the vertex is (-1, -2)) and so it would have to be below the x-axis for all values of r between the two x-intercepts. Setting the expression equal to zero and using the quadratic formula we get \(r=-1 \pm \sqrt2\). Since r cannot be negative, we must choose \(0 and thus \(1 .
I tested the result on a few values of x in the above interval and here is the result:
\(\sqrt2\) is approximately 1.414213562373, and I purposely entered some values of x larger than square root of 2 into the table; these are in the shaded rows. Notice that even if x is very slightly larger, say by one unit in the fifth digit after the point (see the last row) the relationship ( FLOOR(x))^2 = FLOOR(X^2) does not hold
+121 A number of errors crept in when the solution was published;
line 1: "the decimal part of x , \(0" should read "the decimal part of x, \(0 "
line 10 "Since r cannot be negative, we must choose \(0 and thus \(1 . " should read " since r cannot be negative, we must choose
\(0 and thus \(1 "
