Hey, I just have two questions, one of which had been posted on the forum previously. I'm a bit confused about the difference between the two solutions of 32b and 32c if done algebraically. I read and understood the solution for 32b but I'm not sure what the difference the new domain would create
for c) x will have to be smaller than y, right, so if we compute it out, wouldn't it still give up the same algebraic solution as b) did? or would the signs be flipped as we are considering it from the negative side?
I also tried to graphically solve it and have attached it. Could you see if I'm on the right track?
My second question is also attached. I'm just a bit unsure of how it works though I get why theta=0 is a solution ofc.
Firstly note that your rearranged condition is off. You're right about that \(y^2 < x^2 + 1\), but this quadratic inequality solves to give \( \boxed{-\sqrt{x^2-1} < y < \sqrt{x^2+1}} \). On Desmos, you can just type the original condition \(x^2-y^2 < 1\) and the correct plot will still show.
Anyway. The order quantifiers mean everything here. Whilst it turns out both statements are true, the positioning of the quantifiers mean that they mean different things.
The first says that
for all real numbers \(x\)
there exists some \(y\) where \( y > x\), such that \(y^2-x^2 < 1\).
This means that \(x\) has to be any arbitrary real number. And if the statement is true, we should be able to pick some \(y\) (very likely in terms of \(x\)) that will satisfy the condition \(y^2 - x^2 < 1\).
The second says that
there exists a real number \(y\) (not necessarily unique), such that
for all \(x\) where \(x < y\), such that \(y^2-x^2 <1\).
This means that this time, we need to choose some real number \(y\) (and it can't be in terms of anything). That \(y\),
then has to guarantee us that
regardless of what \(x\) is, the condition \(y^2-x^2<1\) is satisfied.
So for the first one, you have to start by assuming \(x\) is whatever real number you want it to be. Then, you need to choose some \(y\in (x,\infty)\) that will work. One example that will work is \( y = \sqrt{x^2 + \frac12} \) - not sure what the answers they give is but that's an example of one of them.
(On Desmos, you can try plotting the required condition \(y>x\), the condition that we want i.e. \( y ^2<x^2+1\), and the curve \(y = \sqrt{x^2+\frac12}\). You'll notice that the curve lies completely between where the two regions intersect, thus affirming this answer is valid.)
Whereas for the second one, you need to commence by choosing some \(y\) that is guaranteed to
always ensure that if \(x < y\), then \(y^2-x^2 < 1\). \(y=0\) should be an intuitively obvious answer here, because then \(y^2 - x^2 = -x^2\), which is always less than 1 regardless of what \(x\) is. (The condition that \(x\in (-\infty,y)\) coincidentally happens to not be that important, because it actually holds for all \(x\in \mathbb{R}\) anyway.)
(This time, on Desmos, to affirm this result we have to plot the line \(y=0\). What we want to check is that no matter what \(x\) is, the condition \(y^2-x^2 < 1\) is satisfied. Converting this graphically, we want to therefore check that the line \(y=0\), is
always inside the region \(y^2-x^2 < 1\) (and doesn't, say, suddenly pop out of it at some value of \(x\). You can check that this will indeed be the case.)
Note that the condition is always \(x < y\) - this stays the same more or less as a consequence of how the questions were written. But the approaches differ because of how the quantifiers are provided.
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With the other question, you can just note that \(y=\sin \theta\) is contained entirely in the 1st and 3rd quadrants in that domain, whilst \(y = m\theta\) is contained entirely in the 2nd and 4th quadrants under the condition that \(m < 0\). Also note that whilst \(\sin \theta = 0\) when \(\theta = -\pi, 0, \pi\), on the other hand \(m\theta = 0\) only when \(\theta = 0\).
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