0
$\begingroup$

I am having trouble with a distance formula proof for my linear algebra class.

This is the proof I have written down:

Verify d(v1,v2) < d(v1,v) + d(v,v2)

d(v1,v2) = ||v1-v2|| = ||v1-0-v2|| = ||v1-v+v-v2|| = ||(v1-v)+(v-v2)|| = ||v1-v|| + ||v-v2|| = d(v1,v) + d(v,v2)

The thing that is confusing me is what is "v"??? The values of v1 and v2 are given I'm guessing, but where does the plain "v" come from? Is it just some arbitrary vector with any value. Because I've been trying to calculate these and they aren't coming out equal. Please help... and an example with numbers would be fantastic.

  • 0
    Can you please tell us a little more about the question? As it stands now, the statement you want to prove is **false**, since it is *not* true in general that $d(v_1,v_2)=d(v_1, v) + d(v_2, v)$. This is why I currently downvoted the question and voted to close it, but I will gladly remove the downvote once you clarify the question. – 2017-02-23
  • 0
    @5xum Ah my mistake, it is actually supposed to be less than or equal to, not equals... I edited it. Does that help? And could you tell me what "v" is? – 2017-02-23
  • 0
    I assume that $d(x,y)$ is given as $\|x-y\|$? In that case, if you want to prove the triangle inequality, then you have to prove that the inequality is true for **all** possible vectors $v_1,v_2,v$. – 2017-02-23
  • 0
    @5xum Okay, so is "v" just any possible vector? And is the above proof now correct? – 2017-02-23
  • 0
    I don't know exactly what $v$ is, since all I know is what you wrote down. It would help if you actually wrote down the entire question that is bothering you, as it's possible there is some confusion in your understanding. But yeah, generally, if a question asks you to prove something in which a free variable exists, then you have to prove it for all possible values of that variable. – 2017-02-23

0 Answers 0