Values of $x$ satisfying $\lfloor x\rfloor + 3\{x\} > 1$ Which values of $x$ satisfy $\lfloor x\rfloor + 3\{x\} > 1$?
$\lfloor \cdot \rfloor$ denotes the greatest integer function and $\{\cdot\}$ denotes the fractional part function.
 A: If $\frac 23 < \{x\} < 1$ then $2 \le 3\{x\} < 3$ and $[x]+2 < [x] + 3\{x\} < [x] + 3$  so  if $[x] \le -2$ this fails but if $[x] \ge -1$ this succeeds.
If $\{x\} =\frac 23$ then $\{x\} = 2$ and $[x] + 3\{x\} = [x]+2 >1$ so if $[x] \le -2$ this fails, and if $x = -1$ this fails (we have $[x]+3\{x\} = 1\not > 1$). But if $x \ge 0$ it succeeds.
If $\frac 13 < \{x\} < 1$ then $1 \le 3\{x\} < 2$ and $[x]+1 < [x] +3\{x\} < [x]+2$. so if $[x]\le -1$ then this fails but if $[x] \ge 0$ this succeeds.
If $\{x\} =\frac 13$ then $[x]+3\{x\} = [x]+1 > 1$ so if $[x]\le 0$ this fails but succeeds if $[x] \ge 1$
If $0 < \{x\} < \frac 13$ then $0 < 3\{x\} < 1$ and $[x] < [x] + 3\{x\} < [x]+ 1$ so this fails if $[x]\le 0$ but succeeds if $[x] \ge 1$.
If $\{x\} 0=$ then $[x]+3\{x\} = [x]$ and this succeeds if $[x] > 1$ but fails if $[x] \le 1$.
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So this fails if $x < -1$ (because the $[x]\le -2$).
If $-1 \le x < 0$ then this succeeds if $\frac 23< \{x\} < 1$, but fails other wise.  So if $x\le -\frac 13$ we have failure but if $-\frac 13< x < 0$ we have success.
If $0 \le x < 1$ we have success if $\frac 13 < \{x\} < 1$, but fail otherwise.  So if $\frac 13 < x< 1$ we have success but $0\le x \le \frac 13$ wil have failure.
If $1 \le x $ then we have failure if $x=1$ but success other wise.
So the solution set is $(-\frac 13,0)\cup (\frac 13,1)\cup (1,\infty)$.
