If $a,b\in R$ are distinct numbers satisfying $|a-1|+|b-1|=|a|+|b|=|a+1|+|b+1|$,then find the minimum value of $|a-b|$

Question

If $a,b\in R$ are distinct numbers satisfying $|a-1|+|b-1|=|a|+|b|=|a+1|+|b+1|$,then find the minimum value of $|a-b|$

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I squared both sides of $|a-1|+|b-1|=|a|+|b|$ and $|a|+|b|=|a+1|+|b+1|$ and equated them but the equation obtained is messy and i am stuck.Please help.

Answer 1

Go through the cases. W.l.o.g. let $a \ge b$.

1. $a>1$ and $b>1$ gives $a+b-2= a+b =a+b+2$ which is impossible.

2. $a<-1$ and $b<-1$: same as case 1

3. $a>1$ and $1>b>0$ gives $a-b= a+b =a+b+2$ which is impossible.

4. $a>1$ and $0>b>-1$ gives $a-b= a-b =a+b+2$ which is impossible.

5. $a\ge1$ and $-1\ge b$ gives $a-b= a-b =a-b$ which is the first possible case. Indeed all numbers satisfying this condition are possible, so the minimum value is $|a-b| = 2$.

The other cases are again impossible.

Answer 2

First of all let's restrict the possible values that $a$ and $b$ can actually take.

• If $a$ and $b$ are strictly positive, then $$a+b+2=|a+1|+|b+1|=|a|+|b|=a+b $$ is never satisfied. Similarly, if $a$ and $b$ are strictly negative, $$|a-1|+|b-1|=|a|+|b|$$ fails.

• It is also easy to check that either $a=0$ or $b=0$ don't work.

• Finally, note that your conditions stay the same if you change sign to $a$ and $b$ at the same time.

For all the reasons above, we shall assume $a>0$ and $b<0$. We can restrict further the possible values that $a$ and $b$ can take. We have $$a-b = |a|+|b| = |a-1|+|b-1| = |a-1| -b+1,$$ since $b-1$ is negative. But this implies $$|a-1| = a-1,$$ which means $a \ge 1$.

Similarly, from $$a-b = |a|+|b| = |a+1|+|b+1| = a+1 +|b+1|$$ we obtain $b \le -1$.

Thus we can finally solve your problem, since $$|a-b| = a-b$$ takes minimum value $2$ as $a$ can't be smaller than $1$ and $b$ can't be bigger than $-1$.

Answer 3

Consider the $16$ cases:

$\hspace{3cm}$

$$\begin{align}1) \ &a,b\ge 1 \Rightarrow \\ & \ \ \ \ \ a-1+b-1=a+b=a+1+b+1 \Rightarrow \emptyset;\\ 2) \ &0\le a\le 1, b\ge 1 \Rightarrow \\ & \ \ \ \ \ -a+1+b-1=a+b=a+1+b+1 \Rightarrow \emptyset;\\ 3) \ &-1\le a\le 0, b\ge 1 \Rightarrow \\ & \ \ \ \ \ -a+1+b-1=-a+b=a+1+b+1 \Rightarrow \\ & \ \ \ \ \ a=-1 \Rightarrow b=1 \Rightarrow (a,b)=(-1,1) \Rightarrow \\ & \ \ \ \ \ |a-b|=2 \ \text{(min)};\\ 4, 7, 8) \ &(a,b)=(-1,1);\\ 5, 6, 11, 12,15,16) \ &\emptyset;\\ 9,10,13,14) \ &(a,b)=(1,-1). \end{align}$$

Answer 4

Let $r=\lvert a-1\rvert+\lvert b-1\rvert=\lvert a\rvert+\lvert b\rvert=\lvert a+1\rvert+\lvert b+1\rvert$. Then $(a,b)$ belongs to three squares simultaneously:

• the square whose vertices are $(\pm r,0)$ and $(0,\pm r)$;
• the square whose vertices are $\bigl(\pm(r+1),0\bigr)$ and $\bigl(0,\pm(r+1)\bigr)$;
• the square whose vertices are $\bigl(\pm(r-1),0\bigr)$ and $\bigl(0,\pm(r-1)\bigr)$.

These squares (when $r=\frac32$) can be seen here:

It is now easy to see geometrically that when $r<2$ (as in the picture) there is no point belonging to all three squares and that when $r=2$ there will be two (and only two) distinct points belonging to all three squares: $\pm(1,-1)$. So, if $(a,b)=\pm(1,-1)$, then $(a,b)$ belongs to all of them and $\lvert a-b\rvert=2$. If $r>2$, then there are other solutions $(a,b)$, bu $\lvert a-b\rvert>2$ then. So, the answer is $2$.