Editing RootTwoIsIrrational
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Most, but perhaps not all, of us know that the square root of 2 can never be written exactly as a fraction. In other words, * EQN:\sqrt{2} is never a ratio of whole numbers, ** EQN:\sqrt{2} is not a rational number, *** EQN:\sqrt{2} is *irrational.* How can we show this? Here are four proofs. ---- !! Algebra The usual way is as follows. * Suppose EQN:\sqrt{2} can be written as a fraction: ** EQN:\sqrt{2}=a/b * Then EQN:2=a^2/b^2 * Then EQN:2b^2=a^2 * Then EQN:a^2 is even ** Hence EQN:a is even *** (This takes a small proof of its own.) * So we can write /a=2k/ * Hence EQN:2b^2=(2k)^2=4k^2 * So EQN:b^2=2k^2 * So now EQN:b^2 is even ** Hence EQN:b is even * So now we can reduce EQN:\frac{a}{b} to a smaller fraction. * Lather, rinse, repeat, as they say. * Following this process we can get smaller and smaller fractions, each of which is exactly the square root of two. Clearly this is nonsense. * The argument is sound, so the premise must be false. * Thus we can never write EQN:\sqrt{2} as EQN:\frac{a}{b} ---- !! Geometry Suppose we can draw a square that has integral sides and an integral diagonal. Using Pythagoras' theorem we can see that this is equivalent to saying that EQN:\sqrt{2} is rational. Consider the following diagram: |>> COLUMN_START [[[ IMG:Sqrt2NotInQ.png ]]] COLUMN_END <<| By Pythagoras we know that if the sides are EQN:a and the diagonal is EQN:b then EQN:2a^2=b^2. This diagram shows that given such a pair, EQN:a,b, we can find a smaller pair. (Exercise for the reader: show that EQN:a and EQN:b in the smaller pair are still integers.) Repeat. Eventually we run out of numbers and thus complete our proof by contradiction ---- !! Algebra, again [[[>50 Exercises for the diligent reader: * Show that EQN:(2a-b)^2=2(b-a)^2 * Show that /b-a/</a/ * what's the relationship between this proof and the previous geometrical one? ]]] Suppose EQN:b^2=2a^2 where EQN:a,b are integers and as small as possible. Then a little algebra shows that EQN:(2a-b)^2=2(b-a)^2 so that EQN:(b-a,2a-b) is a new pair of integers with the same property. But EQN:b-a\lt~a and so we've got a smaller pair of integers whose ratio is EQN:\sqrt{2}, contrary to our original decision to take the /smallest/ pair. Contradiction. ---- !! Continued Fractions [[[>50 Another exercise for the diligent reader: Show that for every number of the form EQN:a/b, its continued fraction will terminate. ]]] Every rational number has a finite continued fraction. We show now that the continued fraction for EQN:1+\sqrt{2} does not terminate. That means EQN:1+\sqrt{2} is irrational, and hence EQN:\sqrt{2} is irrational. To compute the continued fraction we separate the target number into its integer part and the fractional part * EQN:T=1+\sqrt{2}=2+\epsilon where EQN:\epsilon=\sqrt{2}-1 EQN:\epsilon is between 0 and 1 as required, so we can take its reciprocal, giving EQN:\frac{1}{\sqrt{2}-1} which after rationalising the denominator is EQN:1+\sqrt{2}. That's what we started with, so the continued fraction for EQN:1+\sqrt{2} is: * EQN:\LARGE{1+\sqrt{2}=2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+...}}}}} Thus the continued fraction for EQN:1+\sqrt{2} is [2;2,2,2,2,...] which does not terminate. Hence EQN:1+\sqrt{2} is irrational, and so EQN:\sqrt{2} is irrational. ---- CategoryMaths