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Fibonacci's Liber Quadratorum - Proposition 18

```Date: 04/07/2002 at 15:04:03
From: Terri
Subject: Fibonacci's Liber Quadratorum - Proposition 18

Hello,

I've been struggling over a proof by contradiction of Fibonacci's
Proposition 18.  It states that

"If any two numbers (positive integers) have an even sum, then the
ratio of their sum to their difference will not be the same as the
ratio of the larger number to the smaller."

What I need to do is prove this by contradiction, assume that it is
true for m and n, and show that this contradicts the irrationality of
the square root of 2.

Any help would be greatly appreciated. Thank you.
```

```
Date: 04/07/2002 at 17:48:45
From: Doctor Jubal
Subject: Re: Fibonacci's Liber Quadratorum - Proposition 18

Hi Terri,

Thanks for writing Dr. Math.

Well, assuming m and n exist (I'm letting m be the larger), then

m+n     m
----- = ---
m-n     n

n (m+n) = m (m-n)

n^2 + mn = m^2 - mn

2mn = m^2 - n^2

But recall the algebra identity that m^2 - n^2 = (m+n)(m-n)

2mn = (m+n)(m-n)

Now, at this point, note that if you divide both sides by (m-n)^2, the
right-hand side becomes (m+n)/(m-n), which was part of the original
relation we were trying to prove or disprove. This is interesting, and
gives us a relation equivalent to the one we started with.

2mn      m
------- = ---
(m-n)^2    n

Solve this for 2 in terms of m and n, and see if inspiration strikes
you.

Does this help?  Write back if you'd like to talk about this some
more, or if you have any other questions.

- Doctor Jubal, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 04/07/2002 at 19:07:43
From: Terri
Subject: Fibonacci

Hello again. I greatly appreciate your feedback on my question; thank
you. I reached the equation: the square root of 2 = (m-n)/n. So in
order to show that this contradicts the irrationality of the square
root of 2, do I just need to come up with a counterexample?

Thanks again and I hope to hear back from you.
```

```
Date: 04/08/2002 at 09:42:37
From: Doctor Jubal
Subject: Re: Fibonacci

Hi Terri,

Thanks for writing back.

No, you've just set up the contradiction right there. The square root
of two is (m-n)/n. But m and n are integers, so (m-n)/n is a rational
number, which the square root of two isn't. Since the existence of m
and n leads to an impossibility, they don't exist.

Write back if you'd like to talk about this more, or if you have any
other questions.

- Doctor Jubal, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Number Theory
High School Number Theory

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