User:Maelin/Sandbox

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  1. Consider a point A in the plane R2, that is \textstyle A = (x,y) \quad \mbox{where} \quad x,y \in \mathbb{R}
  2. Draw a point B at unit distance from A, such that the line AB forms an angle θ to the x-axis. Now \textstyle B = \left( x + \cos{\theta}, y + \sin{\theta} \right)
  3. Draw a third point C at unit distance from both A and B. Without loss of generality we can assume that C is counterclockwise from B if looking from A. ABC now forms an equilateral triangle, so the line AC forms an angle of θ + π/3 to the x-axis. Now \textstyle C = (x + \cos{( \theta + { \pi \over 3} )}, y + \sin{( \theta + { \pi \over 3} )} )
  4. By the compound angle formula, we have
    \begin{array}{ll} C & = \left( x + \cos{\theta} \cos{{\pi \over 3}} - \sin{\theta} \sin{{\pi \over 3}} \ , \ y + \sin{\theta} \cos{{\pi \over 3}} + \cos{\theta} \sin{{\pi \over 3}} \right) \\ & = \left( x + {1 \over 2} \cos{\theta} - { \sqrt{2} \over 2} \sin{\theta} \ , \ y + {1 \over 2} \sin{\theta} + { \sqrt{2} \over 2} \cos{\theta} \right) \\ \end{array}
  5. We observe that any rational number plus any irrational number is irrational. Proof:
    1. Suppose there exists a rational number a/b, such that a/b = p/q + x, for some integers p and q, and some irrational x
    2. Then, a/b - p/q = x
    3. So (aq - pb) / bq = x
    4. But this means x must be rational. A contradiction!
    5. Therefore, no rational number and irrational number have a rational sum.
  6. In a very similar fashion, we observe that any rational number multiplied by an irrational number is irrational. Proof:
    1. Suppose there exists a rational number a/b, such that a/b = px/q, for some integers p and q, and some irrational x
    2. Then, aq / bp = x
    3. But this means x must be rational. A contradiction!
    4. Therefore, no rational number and irrational number have a rational product.
  7. Now we show that not all 3 of A, B and C can lie in the rational plane. Proof:
    1. Suppose A lies in the rational plane, Q2. x and y must be rational.
    2. Suppose B lies in the rational plane. x + cos(θ) and y + sin(θ) must be rational.
    3. Since no rational number is the sum of a rational and an irrational, cos(θ) and sin(θ) must therefore be rational.
    4. Suppose finally that C lies in the rational plane. \textstyle x + {1 \over 2} \cos{\theta} - {\sqrt{2} \over 2} \sin{\theta} and \textstyle y + {1 \over 2} \sin{\theta} + {\sqrt{2} \over 2} \cos{\theta} must therefore be rational.
    5. Since the rational numbers form a field, they are closed under addition and multiplication. So let X = \textstyle x + {1 \over 2} \cos{\theta} and \textstyle Y = y + {1 \over 2} \sin{\theta}. Now X and Y are rational. Also, \textstyle {1 \over 2} \cos{\theta} and \textstyle {{-1} \over 2} \sin{\theta} are rational.
    6. Note now that \textstyle \sqrt{2} is irrational. So let \textstyle X' = {\sqrt{2} \over 2} \cos{\theta} and \textstyle Y' = {{-\sqrt{2}} \over 2} \sin{\theta}. Since the product of a rational and an irrational is irrational, so must X’ and Y’ be irrational.
    7. Now, we have C = (X + X’, Y + Y’)
    8. But the sum of a rational and an irrational is irrational. So C has irrational co-ordinates. This contradicts our supposition that C lies in the rational plane.
    9. So not all three of A, B and C can lie in the rational plane.
  8. So no equilateral unit triangle exists in the rational plane.
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