Define two point $(a, b)$ and $(c, d)$ to be equal if and only if $b^2+a^2=d^2+c^2$. The task at hand was to prove this an equivalence relation, and to describe and sketch its equivalence classes. I am a poor spatial thinker so I'm struggling to understand how to do the latter. Some help would be greatly appreciated!
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The points are obviously in $R^2$ – Liam Cooney Sep 11 '16 at 03:06
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You dont need spatial thinking for it – Zelos Malum Sep 11 '16 at 03:23
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The individual below has helped me with that. So would the nicest way to write the equivalence class of a point just be to say it's the set of (x,y) in the plane such that the square root of the sum of the squares x and y is the square root of the of the sum of the squares a and b? – Liam Cooney Sep 11 '16 at 03:27
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Sorry for all the words. I format terribly on here – Liam Cooney Sep 11 '16 at 03:27
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If you can wait an hour i will write a non-spatial answer – Zelos Malum Sep 11 '16 at 03:29
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I mean I already understand the individuals answer below and am actually pretty embarrassed I didn't see it, so I can certainly do this part of my homework now but I'd certainly be interested in seeing how you're thinking of it if you're interested in showing me :) – Liam Cooney Sep 11 '16 at 03:31
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I will once i am home, it will be more "proper" and no geometry – Zelos Malum Sep 11 '16 at 04:13
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Sounds good. Thanks! – Liam Cooney Sep 11 '16 at 04:26
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Hows that for an altwrnative? – Zelos Malum Sep 12 '16 at 02:37
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Quite interesting, thank you! – Liam Cooney Sep 16 '16 at 09:30
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I like it more, no geometry only algebra :-) – Zelos Malum Sep 16 '16 at 09:38
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I'm an algebra guy too :) – Liam Cooney Sep 16 '16 at 09:43
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Hint:
$a^2 + b^2 = c^2 + d^2$ if and only if $\sqrt{a^2 + b^2} = \sqrt{c^2 + d^2}$.
Now, $\sqrt{a^2 + b^2}$ is the distance of the point $(a,b)$ from $(0,0)$, and $\sqrt{c^2 + d^2}$ is the distance of the point $(c,d)$ from $(0,0)$.
So $(a,b)$ is equivalent to $(c,d)$ means what?
Caleb Stanford
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So it's the set of points equidistant from the origin...i.e. a circle?...again not a spatial thinker so gotta make sure I've got this straight in my head haha – Liam Cooney Sep 11 '16 at 03:13
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No problem thank you. Would the easiest way to describe the equivalence class of a point (a, b) just be to say it's the set of points (x, y) in the plane such that the square root of the sum of the squares x and y is equal to the square root of the sum of the squares a and b? – Liam Cooney Sep 11 '16 at 03:36
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I sympathize with you, I hate spatial reasoning and avoid it like the plague so here is my approch. As the former is fairly trivial and I assume by your post you have already done it I will go on the latter entirely. Let us write $\sim$ for the relation and we have that $(a,b)\sim(c,d)$ if and only if $a^2+b^2=c^2+d^2$. Notice that the collection of equivalence classes is written as $\mathbb{R}^2/\sim$
Proposition: There exists a bijection $\varphi:\mathbb{R}^2/\sim\to\mathbb{R}_{0\leq}$.
We define $\varphi:\mathbb{R}^2/\sim\to\mathbb{R}_{0\leq}$ by $\varphi((a,b))=a^2+b^2$. We need to check that this is well-defined for completeness, let $(a,b)\sim(c,d)$ and then we have $$\varphi((a,b))=a^2+b^2=c^2+d^2=\varphi((c,d))$$ We will check if it is surjective injective. For surjectivity, let $(a,b)$ be given and $\varphi((a,b))=a^2+b^2=\epsilon$. Then we pick $|a|\leq\sqrt{\epsilon}$ and set $b=\pm\sqrt{\epsilon-a^2}$ and it is satisfied. Injectivity is fairly obvious by the definitions so it is clearly a bijection.
That means we have an inverse and we can write it as $\sigma(r)=\{(\pm a,\pm\sqrt{r-a^2})\in\mathbb{R}^2:a\leq\sqrt{r}\}$ This will give us all equivalence classes, of course any equivalent writing would be just as much correct.
Personally I prefer just prefer, as we are not talking about any specific structure, that the equivalence classes are equivalent to the non-negative real numbers.
Zelos Malum
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