Number theory - greatest integer function

Greatest Integer Function properties:
1. x-1 < [x] <= x < [x] + 1
2. for any integer 'n', [n+x] = n + [x]
3. [-x]
3a.  [-x] = -[x] - 1 if x is not integer
3b. [-x] = -[x] if x is integer
4. [x + y] >= [x] + [y]
4a. In general: [x + y + z ... -w] >= [x] + [y] + [z] -[w]
5. [xy] >= [x][y] and generalize like above


Ex1:
Solve for x.
2[x] = x + 2{x}
Where [] => greatest integer function and {} = fractional part function.
Solution:
Let x = p + q where p = [x] and q = {x} where 0 <= q < 1.
2p = p + q + 2q = p + 3q => p = 3q
Only 3 values of q can give an integer => 
q = 0 => p = 0
q = 1/3 => p =1
q = 2/3 => p = 2
So 3 solutions:
0+0, 1 + 1/3, 2 + 2/3
x = 0,4/3,8/3

Ex2:
36{x}^2 = 6[x]x
Solution:
Let x = p + q where p = [x] and q = {x} where 0 <= q < 1.
=> 36q^2 = 6p(p + q)
=> 6q^2 = p^2 + pq
=> p^2 + pq - 6q^2 = 0
Factorize to get:
p^2 + 3pq - 2pq - 6q^2 = 0
p(p + 3q) -2q(p + 3q) = 0
=> (p - 2q)(p + 3q) =0
p = 2q or p = -3q

Case 1:
p = 2q
Since q is positive:
q = 0 => p = 0
q = 1/2 => p = 1
are the only 2 solutions here.

Case 2:
p = -3q
q = 0 => p = 0
q = 1/3 => p = -1
q = 2/3 => p = -2

Finally:
x = 0+0,  1+1/2, -1+1/3,-2+2/3 = 0, 3/2, -2/3, -4/3 => 4 solutions.

Ex3.
Find number of real solutions for:
[x/2] + [2x/3] = x
Solution:

Let

$$\bigl\lfloor \tfrac{x}{2}\bigr\rfloor+\bigl\lfloor \tfrac{2x}{3}\bigr\rfloor=x, \qquad x\in\mathbb R .$$

1. $x$ must be an integer

Both floor-values are integers, so their sum is an integer; hence $x$ itself must be an integer.

2. Reduce the problem modulo 6

Write every integer as

$$x = 6k+r, \qquad k\in\mathbb Z,\; r\in\{0,1,2,3,4,5\}.$$

Compute the two floors:

$$\left\lfloor\frac{x}{2}\right\rfloor = \left\lfloor\frac{6k+r}{2}\right\rfloor = 3k+\left\lfloor\frac{r}{2}\right\rfloor,$$ $$\left\lfloor\frac{2x}{3}\right\rfloor = \left\lfloor\frac{12k+2r}{3}\right\rfloor = 4k+\left\lfloor\frac{2r}{3}\right\rfloor .$$

Adding and setting equal to $x$:

$$7k+\left\lfloor\frac{r}{2}\right\rfloor+\left\lfloor\frac{2r}{3}\right\rfloor = 6k+r \;\Longrightarrow\; k+\left\lfloor\frac{r}{2}\right\rfloor+\left\lfloor\frac{2r}{3}\right\rfloor = r.$$

3. Solve for each remainder $r$

$r$	$\lfloor r/2\rfloor$	$\lfloor 2r/3\rfloor$	Required $k=r-\lfloor r/2\rfloor-\lfloor 2r/3\rfloor$
0	0	0	0
1	0	0	1
2	1	1	0
3	1	2	0
4	2	2	0
5	2	3	0

4. Construct the solutions

Insert each admissible $(k,r)$ into $x=6k+r$:

• $r=0,\;k=0 \;\Rightarrow\; x=0$

• $r=1,\;k=1 \;\Rightarrow\; x=7$

• $r=2,\;k=0 \;\Rightarrow\; x=2$

• $r=3,\;k=0 \;\Rightarrow\; x=3$

• $r=4,\;k=0 \;\Rightarrow\; x=4$

• $r=5,\;k=0 \;\Rightarrow\; x=5$

No negative $k$ satisfies the table, so there are no negative solutions.

5. Verification

Each candidate satisfies the original equation (quick check: $2\to 1+1=2$, $7\to 3+4=7$, etc.).

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$$\boxed{{\displaystyle \text{Number of real solutions}=6,\text{ namely }x\in \{0,2,3,4,5,7\}\mathrm{.}}}$$