# An irrational "aha!" puzzle

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Brute force or computer simulations are often brought to bear on mathematical problems. Among the motivation for using these approaches is they can obviate the need for careful thought. But careful thought often shows a simpler path to the answer. Occasionally, by argument from analogy to already solved problems, or invoking properties of the type of expression, for example, that the answer must possess, a solution can be found without seemingly solving the puzzle at all. Or, if an equation must be solved, it's one much simpler than first thought. Some people refer to these solutions as coming to them in what they call "Aha!" moments.

I will post a series of "Aha!" puzzles, starting with a rather easy one.

The Golden Ratio g has a number of definitions. The one that provides the best clue to our answer is to note that a rectangle of dimensions g x 1 has the property that if it is cut into a square and another rectangle, the second rectangle has the same proportions as the first.

Prove that g is irrational.

I know its usual symbol is Greek phi; but I don't know how to make one here.

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When a rectangle of dimensions g x 1 is cut into a square and another triangle, the square will be of dimensions 1 x 1 and another rectangle (g - 1) x 1

The proportions of the original rectangle = longer side g / shorter side '1'
The proportions of the second rectangle = longer side 1 / shorter side (g -1), Then
g/1 = 1/(g - 1) which can be written as g = squareroot of (g + 1)

Therefore g is an irrational number.

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When a rectangle of dimensions g x 1 is cut into a square and another triangle, the square will be of dimensions 1 x 1 and another rectangle (g - 1) x 1

The proportions of the original rectangle = longer side g / shorter side '1'

The proportions of the second rectangle = longer side 1 / shorter side (g -1), Then

g/1 = 1/(g - 1) which can be written as g = squareroot of (g + 1)

Therefore g is an irrational number.

Perfect. That is a proof. And even tho not all square roots are irrational this one is.

Now, I'm looking for a proof that begins with the assumption that g is rational and show that it leads to an impossible result.

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g
_________
|            |     |
|         g' |     |
|            |     |
|______|__ |
g'      1

so we have (g'+1)/g' = g'/1

let's assume g is rational.

multiplying g' by both sides we have..

g'+1 = (g')^2

dividing both sides by g'+1 we have...

1 = (g')^2/(g' +1)

so, if g is rational, then g = a/b where a and b are whole numbers.

1 = (a/b)^2 /(a/b +1)

1 = a^2/b^2 /((a +b)/b)

1 = a^2/(b*(a+

so, there are 4 possible cases. a even b even, a odd b odd, a even b odd, a odd b even.

lets get the easy ones out of the way. odd + odd = even, so if both a and b are odd, the denominator will be even, the numerator will be odd, and therefore the fraction can't reduce to 1.

even +odd = odd, so if b is even and a is odd, then once again the fraction can't reduce to 1, as well as vice versa.

now we come to the hard case, both even.

if both are even, the fraction can only reduce to 1 if (2*c)*((2*d)+(2*c)) = (2*d)^2.

which is equal to c*(d+c) = d^2

which can only be true if c = d.

but if c = d, then 2*d^2 = d^2

and we have 2 = 1; a contradiction.

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Golden ratio satisfies

g=1/(g-1)

Assume that it is rational, then we can write g=p/q where gcd(p,q)=1

Now we have p/q=1/(p/q-1), after moving some stuff around you end up with p2-pq=q2 -> p2-q2=pq -> (p+q)(p-q)=pq.

p and q are natural numbers, easy to see that p is not 1, let m be some prime divisor of p, since gcd(p,q)=1 then a does not divide q, on the RHS of the equation we have a but on the LHS we have multiplication of two things that do not divide a so LHS cannot divide a

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Let g = p / q, with p and q both positive integers and gcd(p,q) = 1. g satisfies the equation g^2 = g + 1, therefore (p/q)^2 = (p/q) + 1, and (p^2)/q = p + q. Since p and q are integers, (p^2)/q must be an integer, therefore gcd(p,q) must be at least q . This is a contradiction since q cannot be 1 (sqrt(5), a component of g, cannot be expressed as an integer).

Edited by ThunderCloud
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Assume g is rational.

Then there exists a golden rectangle that is n x m where n>m, and g = n/m.
Now remove an mxm square, leaving an m x n-m rectangle with g = m/(n-m)

Thus, g = n/m = m/(n-m), and both numerators and both denominators are (positive) integers.

But also, m<n and (n-m)<m. That is, the numerator and denominator both decreased.

We can continue removing squares and creating new golden rectangles indefinitely.

But that is impossible since positive integers have a lower bound of unity.

Look Ma, no square roots .

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