0
$\begingroup$

what is solution set ?

$$\lfloor x \rfloor +\lfloor 3x \rfloor =0$$


My Try :

$$\lfloor x \rfloor =x-p_x \\\lfloor 3x \rfloor =3x-p_y$$

So we have :

$$4x-(p_x+p_y)=0\\4x=(p_x+p_y)$$

Now since $$0\leq p_x +p_y <2$$ $$0\leq 4x<2\\0\leq x <\frac{1}{2}$$

it is right ?

$\endgroup$
  • $\begingroup$ If $x=.49$, say, then $\lfloor 3x\rfloor = 1$. $\endgroup$ – lulu Nov 8 '17 at 12:33
  • 1
    $\begingroup$ No it's not right. Try a plot. $\endgroup$ – amcalde Nov 8 '17 at 12:34
2
$\begingroup$

Let $x=a+f$ where $a$ is an integer, $0\le f<1$

$$0=\lfloor x \rfloor +\lfloor 3x \rfloor\ge a+3a\implies a\le0$$

But if $a<0, \lfloor x \rfloor +\lfloor 3x \rfloor<0$

So, $a=0$ and we need $3f<1\iff f<?$

$\endgroup$
  • $\begingroup$ Why is this not acceptable? $a<0, \lfloor x \rfloor +\lfloor 3x \rfloor<0$ $\endgroup$ – Almot1960 Nov 8 '17 at 12:39
  • $\begingroup$ @Almot1960, If $-1<a<0,$ $$\lfloor x \rfloor=?$$ $$-3<3a<0,\lfloor 3x \rfloor =?$$ $\endgroup$ – lab bhattacharjee Nov 8 '17 at 12:44
0
$\begingroup$

First of all $0\le x < 1$. This is not so hard to prove. So really you just need to focus on the $\lfloor 3x\rfloor$ piece.

$\endgroup$
0
$\begingroup$

For solving, you can notice that if $x<0$, both the terms in LHS will be negative, and therefore cann't add up to $0$. If $x>0$ then both terms in LHS will be non-negative. Hence, they'll sum up to $0$ only when both the terms are $0$ individually. I.e. $$\lfloor x\rfloor =\lfloor 3x\rfloor =0 \implies x \in [0,1) \cap [0,1/3)$$

$$\implies x \in [0,1/3)$$


What you've done is correct, but you don't get the complete solution set for the given equation, because the inequality used by you isn't "strong" enough.

What you get from this is a 'superset' of the complete solution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.