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Normally I can solve these problems but THIS one irritates me.

$d(x,y)=|x-2y|$ on the real numbers.

I use the axioms (from my course where I'm a student, see Image: http://puu.sh/BfjWu/7f75958f49.png).

M1: $d(x,y)=0\Leftrightarrow x=y$

M2: $d(x,y)=d(y,x)$ (symmetry)

M3: $d(x,y) \leq d(x,z)+d(z,y)$ (triangle inequality)

So can anyone tell me why M2 is wrong? I know M1 fails because we can let $x=1$ and $y=1$ then $1\neq 2$

I know M3 fails because I can let $x=5,z=2,y=1$ then $d(5,1)=|5-2\cdot 1|=3>1+0=|5-2\cdot 2|+|2-2\cdot 1|=d(5,2)+d(2,1)$

But I can't figure out why M2 fails... First I tried this:

$d(x,y)=|x-2y|=|2y-x|=d(y,x)$

Then M2 holds, but I think it's wrong. In the other hand, I see M2 fails if:

$d(x,y)=|x-2y| \neq |y-2x|=d(y,x)$

Can anyone help me to clear it up?

Rodrigo de Azevedo
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Amir Hassan
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  • Try $x = 1$ and $y = 0$. – J126 Aug 16 '18 at 18:45
  • Yea, but let x=1 and y=2 then |1-22|=|22-1| which gives me |-3|=3 and the numerical of -3 is 3, hence 3=3

    Edit: If i let x=1 and y=0, then I will end up with |1-20|=|20-1| which is also gives me 1=1.

     – Amir Hassan Aug 16 '18 at 18:46
  • Btw I use my idea that: $d(x,y)=|x-2y|=|2y-x|=d(y,x)$ but do I really have to use $d(x,y)=|x-2y|=|y-2x|=d(y,x)$ because x and y have swapped places but not the constant 2. If I should use the last one I typed, then I see M2 fails. – Amir Hassan Aug 16 '18 at 18:49
  • $|x-2y| = |-(2y-x)| = |2y-x|$ – Cesareo Aug 16 '18 at 18:56
  • @KasperMedK Your calculation is wrong. If we let $x=1$ and $y=0$, then $\lvert x-2y \rvert =1$, and $\lvert y-2x \rvert = 2$. – Sam Wong Sep 19 '19 at 12:13

1 Answers1

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Yeah! Your last argument is the correct one.

You have to show that |x-2y|=|y-2x|.

Which obviously fails because x=1 &y=0 as given in comments does not satisfy it

Devendra Singh Rana
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