Can you actually have an isosceles right triangle with $45$ degree angles?
Such a triangle has sides of $2$, $\sqrt{2}$, and $\sqrt{2}$. But $\sqrt{2}$ is an irrational number. Can you actually have line segments which are $\sqrt{2}$ in length?
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Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Oct 18 '23 at 10:05
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"Can you actually have line segments which are $\sqrt{2}$ in length?" - yes, why would you think there is a problem? – mihaild Oct 18 '23 at 10:06
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The answer is trivially affirmative, we can easily construct an isosceles right triangle (and consequently it has two $45$ degrees angles).
Proof. $\mathbb{R}$ is a complete set (having a cardinality of $2^{\aleph_0}$) and $\sqrt{2}, 2 \in \mathbb{R}$. Now, we can construct our triangle from the square $A$ of area $2$. Just cut it in half joining any pair of opposite vertices in order to get an isosceles right triangle of unit area. Its cathetes length is $\sqrt{2}$, since the side of the aforementioned square measures $\sqrt{\textit{Area of} A}$, which is $\sqrt{2}$.
Thus, we have shown that if a square of area $2$ exists, then an isosceles right triangle with cathetes of $\sqrt{2}$ exists, but $x^2=2 : x \in \mathbb{R}^+$ has one solution in the given field (i.e., $\mathbb{R}^+$) so that our first statement is proven.
Marco Ripà
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My puzzle: Sq rt 2 is an irrational number. They are described as indeterminate. How can you conceptualize a line segment of indeterminate length? I assume the segment will have definite endpoints, and the length of it is determinate, ie, it has a precise cutoff point on the line. – K Mann Oct 18 '23 at 13:17
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@mihaild I added it in brackets as an additional info, just to say that there is a bijection between any point on a line and one (and only one) element of the set $\mathbb{R}$. – Marco Ripà Oct 18 '23 at 19:11
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@KMann If you can't write the exact value of $\sqrt{2}$ or $\pi$ on a piece of paper it doesn't mean that you cannot define a circle of unit radius... – Marco Ripà Oct 18 '23 at 19:11
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@MarcoRipà just bijection here is not important, topology is important (and actually I think line is usually essentially just defined as $\mathbb R$). – mihaild Oct 18 '23 at 20:04
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@KMann "They are described as indeterminate" - where? Irrational numbers are perfectly normal. – mihaild Oct 18 '23 at 20:04
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