# Equivalence relation with the floor function

Let us consider a function $$f:\ \Bbb{R}\ \longrightarrow\ \Bbb{R}$$ and we define the equivalence relation $$\sim$$ on $$\Bbb{R}$$ such that: $$x\ \sim\ y\qquad\Leftrightarrow\qquad f(x)=f(y).$$

Note: The following $$\lfloor x\rfloor$$ is the floor function of $$x$$. It gives you the first integer smaller or equal to $$x$$. For example $$\lfloor2.2\rfloor=2$$, $$\lfloor12\rfloor=12$$, and $$\lfloor−2.2\rfloor=−3$$.

For the function $$f(x)=\lfloor x/10\rfloor$$, only one of the following statement is true.

For $$x=a_0+10a_1+10^2a_2+...+10^na_n$$, with all $$a_j\in\{0,1,2,\ldots,9\}$$ and for all $$c\in\{0,1,2,\ldots,9\}$$:

1) $$x\ \sim\ x-c$$ if $$a_0-c<5$$,

2) $$x\ \sim\ x+c$$ if $$a_0+c<10$$.

My working: Take $$a_0=9$$ and $$c=2$$ then the first statement is false as $$9-2>5$$ but if I apply to the second statement $$9+2>10$$ which is wrong too, so I think my method is wrong.

I am guessing the second one is true from the function $$f(x)=\lfloor x/10\rfloor$$, but I'm really not sure about this. Any help will be appreciated!

Thanks.

• You seem to suspect that the first is false; have you looked for a counterexample? – Servaes Nov 1 '18 at 17:03
• Assuming $a_0$ represents the "units place" int the standard base $10$ expression for $x$, then your second formula is not correct. Let $x=1.9, c=8.9$. Then $f(x)=0$ but $f(x+c)=f(10.8)=1$ (while $a_0+c=1+8.9=9.9<10$). – lulu Nov 1 '18 at 17:06
• Your first formula is false as well. Let $x=10,c=1$. Or am I misunderstanding the definition of $a_0$? – lulu Nov 1 '18 at 17:09
• I've just done something similar and it suggests that both statements are false so I'm guessing I haven't used a correct method? – BlackSky Nov 1 '18 at 17:20
• Your working is irrelevant to the first statement; you get $9-2>5$ so your example is not within the scope of the first statement. The first statement can be expressed as $$a_0-c<5\quad\Rightarrow\quad x\sim x-c,$$ so to disprove it you need to start with $a_0$ and $c$ such that $a_0-c<5$. – Servaes Nov 1 '18 at 17:21

## 2 Answers

It's just decimal representation. If you subtracta a $$c$$ from a number do you have to carry/borrow a one or not.

to wit....

So $$x = 10^na_n + 10^{n_1}a_{n-1} + ..... + 10a_1 + a_0$$

$$\frac x{10} = 10^{n-1}a_n + 10^{n-2}a_{n-1} + ..... + a_1 + \frac {a_0}{10}$$

$$f(x) = \lfloor \frac x{10} \rfloor = 10^{n-1}a_n + 10^{n-2}a_{n-1} + ..... + a_1$$

And $$x - c = 10^na_n + 10^{n_1}a_{n-1} + ..... + 10a_1 + (a_0-c)$$

And $$\frac {x-c}{10} = 10^{n-1}a_n + 10^{n_2}a_{n-1} + ..... + a_1 + \frac {(a_0-c)}{10}$$

To find the least integer depends upon the value of $$\frac {(a_0-c)}{10}$$.

$$-9 \le a_0 -c \le 9$$ and $$-.9\le \frac {(a_0-c)}{10} \le .9$$

If $$\frac {(a_0-c)}{10} \ge 0$$ then

$$\rfloor \frac {x-c}{10}\lfloor = 10^{n-1}a_n + 10^{n_2}a_{n-1} + ..... + a_1 + \rfloor\frac {(a_0-c)}{10}\lfloor= 10^{n-1}a_n + 10^{n_2}a_{n-1} + ..... + a_1 = f(x)$$

If $$\frac {(a_0-c)}{10} \le 0$$ then

$$\rfloor \frac {x-c}{10}\lfloor = 10^{n-1}a_n + 10^{n_2}a_{n-1} + ..... + a_1 + \rfloor\frac {(a_0-c)}{10}\lfloor= 10^{n-1}a_n + 10^{n_2}a_{n-1} + ..... + a_1 - 1= f(x)-1$$

So $$x$$~$$x-c$$ if $$a_0 -c \ge 0$$. But $$x$$ !~ $$x-c$$ if $$a_0 - c < 0$$. So if $$a_0 - c < 0 < 5$$ this is not true.

....

You can do 2) the same way.

$$f(x+c) = 10^{n-1}a_n + 10^{n-2}a_{n-1} + ..... + a_1 + \lfloor \frac {a_0+c}{10}\rfloor = f(x) + \lfloor \frac {a_0+c}{10}\rfloor$$.

And $$\lfloor \frac {a_0+c}{10}\rfloor = 0$$ if $$a_0+c< 10$$ but $$\lfloor \frac {a_0+c}{10}\rfloor = 1$$ if $$a_0 + c \ge 10$$.

So 2) is true.

The question is a bit of a mess, so let's clean things up; note that if $$x=a_0+10a_1+10^2a_2+\ldots+10^na_n,$$ with $$a_j\in\{0,1,2,\ldots,9\}$$, then $$f(x)=\lfloor\frac{x}{10}\rfloor$$ is simply $$f(x)=a_1+10a_2+10^2a_3+\ldots+10^{n-1}a_n.$$ Now if $$c\in\{0,1,2,\ldots,9\}$$ then $$x\sim x-c$$ if and only if $$x-c=b+10a_1+10^2a_2+\cdots+10^na_n,$$ for some $$b\in\{0,1,2,\ldots,9\}$$. That is, if and only if $$a_0-c\geq0$$. Similarly $$x\sim x+c$$ if and only if $$x+c=b+10a_1+10^2a_2+\cdots+10^na_n,$$ for some $$b'\in\{0,1,2,\ldots,9\}$$. That is, if and only if $$a_0+c<10$$. This is precisely the second statement, so this one is true and the other one seems false.

To show that the first statement is false, find $$a_j$$ and $$c$$ such that $$a_0-c<0$$. Simply take $$x=0$$ and $$c=1$$.Then $$a_0-c=-1<5$$ but $$x\not\sim x-c$$ because $$f(x)=0$$ and $$f(x-c)=-1$$.