For all $x$ and some of $y$

Question

Prove that this works for all $x$ and and only some $y$ $$\sqrt{(x-1)^2-(y+2)^2}=0.$$

This is as far as I got so far

Difference of squares:

$\sqrt{(x-1-y-2)(x-1+y+2)}=0$
$\sqrt{x-y-3}\sqrt{x+y+1}=0$

Therefore $x-y-3=0 \implies y=x-3$

$x+y+1=0$ and $y=-1-x$

I just don't know where to go from here?

Answer 1

What you must mean (and it really helps to write it clearly) is that for all $x$ you can find a $y$ that makes the statement true and you can find another $y$ that makes the statement false. So given an $x$ you can exhibit a $y$ that makes the statement true-just solve for $y$
$$(x-1)^2=(y+2)^2\\y+2=\pm(x-1)\\ y=\begin {cases}x-3\\-1-x \end{cases}$$ Either of these $y$'s will work, which you can show by plugging back in. Now let $y=x-4$, for example, to show that there is at least one $y$ that doesn't work.

Note that if you choose $y$ first, the same thing happens. For all $y$ there are some $x$ that work and some $x$ that do not work.

Answer 2

What you need to show is the following:

For every real number $x$, there exists a real number $y$ such that $$\sqrt{(x-1)^2-(y+2)^2}=0,\tag{Eqn}$$

An equivalent way of saying "there exists a real $y$" is to say "for some real $y$". Saying "for some of $y$" brings to mind other things that aren't relevant. For each $x$, you only need to show that there is (at least) one $y$ satisfying (Eqn).

We write "for all x" and "there is y" using the quantifiers $\forall x$ and $\exists y$. In what follows, both variables $x$ and $y$ are assumed to be real numbers.

Let's rewrite what you have to show using quantifiers: $$ (\forall x)(\exists y)\, \sqrt{(x-1)^2-(y+2)^2}=0,\tag{*} $$ Clearly (yes?), the following statement implies (*): $$ (\forall x)(\exists y)\, [(x-1)^2 = (y+2)^2].\tag{**} $$

So, given $x$, simply take $y=x-3$. Then $x-1 = y+2$, so $(x-1)^2 = (y+2)^2$. This proves (**), which implies (*).

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Your question (but not your title) says "only some", so it seems you also want to show the following:

For every real number $x$, there exists a real number $y$ such that $$\sqrt{(x-1)^2-(y+2)^2}\ne0,\tag{not-Eqn}$$

This is also straightforward. Given $x$, here's how to find a $y$ such that (not-Eqn) is true:

• if $x \ne 1$, let $y=-2$, and
• if $x = 1$, let $y = 0$.

So the second highlighted statement is also true.