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Principles of Mathematical Analysis, Exercise 1.16 (here is a semi-duplicate, same question, but different emphasis):

Suppose $k \ge 3, x,y \in \mathbb{R}^k, | x-y |=d>0,$ and $r>0.$ Prove that if $ 2r>d,$ there are infinitely many $ z \in \mathbb{R}^k$ such that $$| z- x| =| z - y| = r. $$

On the one hand, the proof seems quite intuitive: If $2r > d$, then $x, y$, and $z$, seen as points, form a triangle in $\mathbb{R}^k$. The orbits of $z$ form a circle in the $\mathbb{R}^k$, and obviously the number of $z$ is infinite.

However when one simply takes the equation $| z- x | =| z - y| (= r)$ and computes, it turns out different:

Since $|z- x| =| z - y| $ and for any vector $w\in \mathbb{R}^k(k \geq 3)$ , $| w|^2=w^2$, we have: $$(z-x)^2=(z-y)^2,$$ hence $$z^2-2zx+x^2=z^2-2zy+y^2,$$ $$x^2-y^2=2z(x-y),$$ $$(x+y)(x-y)=2z(x-y),$$ since $x-y \neq 0$ (for $|x-y|>0$), we have $z=(x+y)/2$, which means $z$ is unique rather than infinitely many.

Any idea as to what went wrong above would be appreciated.

C_Arietta_C
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  • What sense are you giving to the square of a vector? Or in general the product of two vectors? – Bruno B Jul 10 '23 at 11:10
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    @BrunoB - It should be the dot product. But then it's not possible to divide by $x-y$. – mr_e_man Jul 10 '23 at 11:16
  • @BrunoB My ideas about vector product are rather murky, but I believe here we are dealing with (copy and paste) the so-called "inner product" (or scalar product) of vectors, and that the distributive law holds well with it. – C_Arietta_C Jul 10 '23 at 11:23
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    $(x+y).(x-y)=2z.(x-y)$ surely has infinitely many solutions. – geetha290krm Jul 10 '23 at 11:23
  • @geetha290krm But can't we just cancel $x-y$? it's nonzero, right? Cancellation law should imho hold here... – C_Arietta_C Jul 10 '23 at 11:33
  • When having vectors , $AX=BX$ , we can not cancel the X : We will get $(A-B)X=0$ & that will have "multiple" answers ! – Prem Jul 10 '23 at 11:35
  • @Prem One way out seems to be viewing $x$ and $y$ as varying themselves, is that what you mean? But then $z$ still lies in the middle point of $xy$, which is not the same with the first proof (and intuition). – C_Arietta_C Jul 10 '23 at 11:37
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    There is no "cancellation law": $(1, a , -1) \cdot (1, 0, 1) = 0 = (0,0,0) \cdot (1,0,1)$ yet $(1,a,-1) \neq (0,0,0)$ for example, giving an infinity of solutions to $x \cdot (1, 0, 1) = 0$ just by making one coordinate vary, and of course there are other solutions. – Bruno B Jul 10 '23 at 11:42
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    @C_Arietta_C it is probably best that you think geometrically. For a fixed nonzero vector $v \in \mathbb{R}^{3}$, the set of vectors $x \in \mathbb{R}^{3}$ for which $x \cdot v = 0$ is a two-dimensional plane with $v$ being orthogonal to it. So it is definitely an infinite set. Now, your equation is not that of a plane, but it can definitely allow for an infinite set of solutions, since you are dealing with scalar products and there is no kind of cancellation rule as explained above. – Diego Marcon Jul 10 '23 at 12:08
  • @DiegoMarcon I see, what an error! Thanks! – C_Arietta_C Jul 10 '23 at 13:40
  • @BrunoB indeed! Thank you for the enlightenment! – C_Arietta_C Jul 10 '23 at 13:41
  • My Earlier Comment can "Equivalently" be put like this : When having vectors like this $(x+y)(x−y)=2z(x−y)$ { where my $X$ is $(x−y)$ , with $A$ being $(x+y)$ & $B$ being $2z$ } we can not Cancel that Common $X$ Part : We can move the terms around to get $(x+y)−2z=0$ which is my $(A−B)X=0$ : That has "multiple" Solutions when using vectors ! – Prem Jul 10 '23 at 14:53

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