1. ## Contradictions when dividing by zero.

Consider the following mathematical demonstration (which is also my signature):

Let $\;\; x=y$

Then:

$x^2 = xy$

$x^2 - y^2 = xy - y^2$

$(x+y)(x-y) = y(x-y)$

$x+y = y$

$2y = y$

$2 = 1$

Okay, I know that the issue here is a division by zero in the 3rd line down, since $(x-y) = 0$. This problem was shown to me in a textbook, and I found it really interesting how the fact that something illogical (division by zero) will force another illogical statement (2 = 1). I was wondering if anybody knew of some other clever, but simple, deomnstration of this type, because these are very interesting to me. Plus, I kinda wanted as many people to post them on here as possible, and then for the most clever yet simple example to be choosen as the winner. So, lets begin. (Also, it can involve calculus, it doesn't have to be super simple. I'm just hoping to not have to start talking about Rings and Fields, or Topological spaces and Isomorphisims in these examples; thats all I mean by "simple")

Post your favorite little algebraic/calculus demonstration of how a seemingly proper mathematical process can end up in a logical contradiction here, please. Also, please don't point out the "catch" (the 'catch' in the demonstration I supplied is the division by zero in line three), I'd like to see how many of them I can figure out on my own, and I'm sure others would like a chance to try and figure it out on their own also. Thanks in advance for any posts.

$\sqrt{ab} = \sqrt{a}\sqrt{b}$

$\sqrt{1} = \sqrt{1}$

$\sqrt{(1)(1)} = \sqrt{(-1)(-1)}$ Since 1 = 1*1 and 1 = (-1)*(-1)

Using the above property:

$\sqrt{1}\sqrt{1} = \sqrt{-1}\sqrt{-1}$

$(1)(1) = (i)(i)$

$1 = i^2$

$1 = -1$

And this is obviously not true.

3. http://www.mathhelpforum.com/math-he...or-147694.html

http://www.mathhelpforum.com/math-he...on-148441.html

Bacterius's example in the latter thread in post #5 involving complex numbers is a good example, I think. I suppose the more intricate and subtle you can make the error, the less my claim in post #4 of that thread

Originally Posted by undefined
It's good to realise though that to people experienced in mathematics, the errors in these types of "proofs" tend to stand out immediately and appear rather silly.
applies. As someone who writes lots of computer programs, I know just how easy it can be to miss some small mistake that can lead to strange results.

By the way, in logic the contradiction "a and not a" implies the truth of any statement, that is

for all q for all a, "a ^ ~a" -> q

So, if a horse is not a horse, then I'm the king of England.

4. Here's one I stumbled upon a few weeks ago. See if you can get it. We define a rational number in the following way:
$x \in \mathbb{Q}$ iff there exist integers $y$ and $z$ with $z \neq 0$ such that $x = y/z$

Let $a$ and $b$ be real numbers with $b \neq 0$. Let the product of $a$ and $b$ be a rational number:
(i) $ab \in \mathbb{Q}$
so that there exist integers $m$ and $n$ with $n \neq 0$ such that
(ii) $ab = m/n$
Multiply both sides by $1/b$ (remember that $b \neq 0$), and get
(iii) $a = m/bn$
Therefore, by the definition of a rational number, $a$ is rational:
(iv) $a \in \mathbb{Q}$
From (i) and disjunctive addition, we get
(v) $a \in \mathbb{Q} \vee b \in \mathbb{Q}$
By De Morgan's Law, this is equivalent to
(vi) $\neg(a \notin \mathbb{Q} \wedge b \notin \mathbb{Q})$

So, by conditional proof, we've derived the following formula
(vii) $(ab \in \mathbb{Q} \wedge b \neq 0) \rightarrow \neg(a \notin \mathbb{Q} \wedge b \notin \mathbb{Q})$
the contrapositive of which is
(viii) $(a \notin \mathbb{Q} \wedge b \notin \mathbb{Q}) \rightarrow \neg(ab \in \mathbb{Q} \wedge b \neq 0)$
Now we can simplify this a bit. Assume the antecedent of the conditional.
(ix) $a \notin \mathbb{Q} \wedge b \notin \mathbb{Q}$
Then modus ponens gives us
(x) $\neg(ab \in \mathbb{Q} \wedge b \neq 0)$
Which is equivalent to
(xi) $ab \notin \mathbb{Q} \vee b = 0$
We know by our initial assumption that $b \neq 0$, so
(xii) $ab \notin \mathbb{Q}$
Thus by conditional proof from our assumption (ix) and (xii)
(xiii) $(a \notin \mathbb{Q} \wedge b \notin \mathbb{Q}) \rightarrow ab \notin \mathbb{Q}$

Obviously, there is a counterexample to this "theorem." Let $a = \sqrt{2}$ and $b = \sqrt{2}$. Then $ab = \sqrt{2}\times\sqrt{2} = 2$, which is rational.

So what's wrong with the argument?

5. Originally Posted by quiney
Let $a$ and $b$ be real numbers with $b \neq 0$. Let the product of $a$ and $b$ be a rational number:
(i) $ab \in \mathbb{Q}$
so that there exist integers $m$ and $n$ with $n \neq 0$ such that
(ii) $ab = m/n$
Multiply both sides by $1/b$ (remember that $b \neq 0$), and get
(iii) $a = m/bn$
Therefore, by the definition of a rational number, $a$ is rational:
(iv) $a \in \mathbb{Q}$
The definition of rational number does not apply; the author magically assumes bn is an integer!

6. Here's one from Infinite Series.
. . Yes, the fallacy is obvious to those in-the-know,
. . .but it's still surprising at first reading.

We have this infinite series:

$\ln(1+x) \;=\;x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \dfrac{x^5}{5} - \dfrac{x^6}{6} + \hdots$

Let $x = 1$

. . $\ln 2 \;=\;1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \hdots$

. . . . . $=\; \left(1 + \frac{1}{3} + \frac{1}{5} + \hdots\right) - \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \hdots \right)$

Add and subtract $\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \hdots\right)$

. . $\ln 2 \;=\;\left(1 + \frac{1}{3} + \frac{1}{5} + \hdots\right) + \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \hdots\right)$
. . . . . . . . . . . $-\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \hdots\right) - \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \hdots\right)$

. . . . . $=\;\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \hdots\right) -2\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \hdots\right)$

. . . . . $=\;\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \hdots\right) - \left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \hdots\right)$

. . . . . $=\;0$

Hence: . $\ln 2 \:=\:0\quad\Rightarrow\quad e^0 = 2$

Therefore: . $1 = 2$

This is one of the very few that does not use division-by-zero.

7. Originally Posted by Soroban
[size=3]Add and subtract $\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \hdots\right)$
Since no one else replied so far... here you add and subtract a divergent sum.

8. You can start with 1=2 to show that the integers are trivial,

If $1=2$ then $0=1-1=2-1=1$ and so for all $n \in \mathbb{Z}$ you have $n=1+\ldots +1 =0+\ldots =0 \Rightarrow \mathbb{Z}=\{0\}$.

I've always quite liked this proof.

9. Also, there is this proof of 1=2 which doesn't use division by zero,

$-2 = -2$
$\Rightarrow 4 - 6 = 1 - 3$
$\Rightarrow 4 - 6 + 9/4 = 1 - 3 + 9/4$
$\Rightarrow (2 - 3/2)^2 = (1 - 3/2)^2$
$\Rightarrow 2 - 3/2 = 1 - 3/2$
$\Rightarrow 2=1$

10. Thank you everybody for your posts, although I was only able to figure out one of the problems on my own, it was still all very interesting. I also came across another one that you guys may or may not have seen before. Here it is:

[1] Consider a ladder of length $L$ standing up against the wall of a building. The ladder, the ground, and the wall of the building each together form a right triangle, with the wall and the ground as the sides, and the ladder as the hypoteneous.

[2] Also consider that a person on the ground level grabs the ladder at the bottom, and pulls out-ward horizontally with a constant velocity of $v$.

[3] Now, let $x$ denote the horizontal distance from the ladder to the wall, at time $t$

[4] Let $y$ denote the vertical distance of the part of the ladder touching the building to the ground, at time $t$

[5] Now, from all the above, the following relationship can be derived:

$x^2 + y^2 = L^2$

[6] From that we can easily get to:

$y = \sqrt{L^2 - x^2}$

[7] Differentiating with respect to $t$, we get:

$\frac{dy}{dt} = - \frac{(x)(\frac{dx}{dt})}{\sqrt{L^2 - x^2}}$

[8] Since the bottom of the ladder is being pulled away with constant velocity we have $\frac{dx}{dt} = v$; which implies:

$\frac{dy}{dt} = - \frac{(x)(v)}{\sqrt{L^2 - x^2}}$

[9] As the ladder is pulled backwards, obviously the vertical distance to the ground is being reduced, actually its seemingly and intuitively going to zero. In other words:

$x \to + \;\; \mathrm{(as\;\; x\;\; increases)} \;\; \Rightarrow y \to 0$

In fact, since the ladder is being pulled away from the wall, its getting closer to the ground. If the ladder were lying on the ground, but touching the wall, then we'd have $x = L$; so, to decribe it's movements more precisely:

$\lim_{x \to L} \; (y) = 0$

Now, in the expression:

$\frac{dy}{dt} = - \frac{(x)(v)}{\sqrt{L^2 - x^2}}$

Allow for the following simplification in notation:

$-(x)(v) = p$

and

$\sqrt{L^2 - x^2} = q$

so that

$\frac{dy}{dt} = \frac{p}{q}$

Now, we have the following limits:

$\lim_{x \to L} \; (p) = -Lv$

and

$\lim_{x \to L} \; (q) = 0$

[10] Therefore, it follows that:

$\lim{x \to L} \frac{dy}{dt} = - \infty$

In other words, the ladder's top end with fall downward (negitive infront of the infinity) infinitely fast as the bottom end comes closer and closer to being the same distance away from the wall $x$ that the ladders length is $L$; like I said, such that $x \to L$.

Phew, that seemed like allot to write for one problem. Anyway, there it is, I found it on the internet earlier today.