Examples of fallacies in arithmetic and/or algebra I'm currently preparing for a talk to be delivered to a general audience, consisting primarily of undergraduate students from diverse majors.  My proposed topic would be Examples of fallacies in arithmetic and/or algebra.
So my question would be:

What are some examples of arithmetic/algebraic fallacies that you know of?

One example per answer please.
Let me give my own example, which is one of my personal favorites:

Let $$a = b.$$
  Multiplying both sides by $a$, we get $$a^2 = ab.$$
  Subtracting $b^2$ from both sides, we obtain $$a^2 - b^2 = ab - b^2.$$
  Factoring both sides, we have $$(a + b)(a - b) = b(a - b).$$
  Dividing both sides by $(a - b)$, $$a + b = b.$$
  Substituting $a = b$ and simplifying, $$b + b = b,$$ and $$2b = b.$$
  Dividing both sides by $b$, $$2 = 1.$$

Of course, this fallacious argument breaks down because we divided by $a - b = 0$, since $a = b$ by assumption, and division by zero is not allowed.
 A: You might want to look at Edward J. Barbeau's books "Mathematical Fallacies, Flaws, and Flimflam" (MAA, 2000) and "More Fallacies, Flaws and Flimflam" (MAA, 2013).
A: Without resorting to $i$:
$$1-3=4-6\\
1-3+\frac94=4-6+\frac94\\
1-2\cdot1\cdot\frac32+\left(\frac32\right)^2=4-2\cdot2\cdot\frac32+\left(\frac32\right)^2\\
\left(1-\frac32\right)^2=\left(2-\frac32\right)^2\\
1-\frac32=2-\frac32\\
1=2$$
A: I have an answer to this question!
Proposition: Every positive number can be described in less than $15$ English words.
Note: I don't necessarily mean the spelling expansion of its form like $256$ as "two hundred and fifty-six" but also as the more economical "sixteen squared".
Base Step: $1$ can be written in less than fifteen English words.
Hypothesis: Let us assume that till a given positive number $k$, we are able to express all the numbers in less than fifteen English words.
Induction Step: Let us consider $k+1$. Either it can be written in less than fifteen English words or it can't. The first case solves our problem. If it can't be expressed in less than $15$ English words, then it is "the smallest positive integer that cannot be expressed in less than fifteen English words", which is a fourteen-word description!
Therefore, $P(k+1)$ is true wherever $P(k)$ is true.
This implies that all positive integers can be expressed in less than fifteen English words!
(However, it is false. The set of positive integers is infinite. The set of English words is not. Therefore a bijective mapping cannot be made. I'll leave it to you to ponder the fallacy.)
A: $$S=1+2+4+8+\cdots$$
$$2S=2+4+8+16+\cdots$$
Subtracting like this:
$$S-2S=(1+2+4+8+\cdots)-(0+2+4+8+\cdots)=(1-0)+(2-2)+(4-4)+\cdots=1$$
$$S=-1$$
A: I am fond of fallacies where a property of members of set of things, and the properties of the limit of that set, are assumed to be equal. But the limit need not be a member of the set, and therefore need have nothing in common with members of the set.
For example, imagine a collection of line segments that goes straight up one unit and straight right one unit. The total length is two.
Now imagine it goes up a half, right a half, up a half, right a half. Again, the length is two. And now we have something that looks like a staircase.
Now up a third, right a third, up a third, right a third, up a third, right a third.  Another staircase. Length is still two.
Obviously as we continue this sequence the line segments more and more closely approximate a line of length root-two going diagonally.  The conclusion we fallaciously reach is that two and its square root are equal.
A: 1 = 0?
Given the progression $a_n=(-1)^n$, the sum $s_n=\sum_{i=1}^\infty a_i$ can be built in two ways, all terms in parentheses are zero.


*

*$$s_n=\sum_{i=1}^\infty a_i=(1-1)+(1-1)+(1-1)+...=0$$

*$$s_n=\sum_{i=1}^\infty a_i=1+(-1+1)+(-1+1)+...=1$$
Therefore, $s_n=1=0$.
A: I like the following one. It's kinda silly, but still interesting.
We know $1\$=100c$. But then:
$$\begin{align}1\$&=100c\\
&=10c\times 10c\\
&=0.1\$\times 0.1\$\\
&=0.01\$\\
&=1c\end{align}$$
So a dollar is worth just a penny!
A: $$1 = \sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{-1}\sqrt{-1} = i\cdot i = -1$$
