+0

# help

0
49
3

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.

Dec 11, 2019

#1
+118
+1

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

Dec 11, 2019
#2
+118
+1

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 "

Dec 11, 2019
#3
+118
+1

Ha!!?? the same distortions happened again. I used latex to write three inequalities involving x and r , but all three were somehow Butchered in the process of uploading. Well I don't think images will be distorted, so here they come as pictures:

Dec 11, 2019