Dodgy Proofs

Consider these dodgy proofs. Although the results are obviously wrong, where, precisely, do the 'proofs' break down? There are four starter questions, four main questions and three questions for pudding. Good luck!
 
 
STARTER QUESTIONS 

S1. A pound equals a penny   

Proof:
£$1 = 100 p = (10p)^2 = ($ £$0.1)^2 =$ £$0.01 = 1p$


S2. ${\mathbf{2 = 3}}$

Proof:
We have $2\times 0 =0$ and also $3\times 0 = 0$. 
Therefore we have $2 \times 0 = 3 \times 0$. 
Dividing both sides by $0$ gives $2=3$.

S3. The perimeter of a square is four times its area

Proof:
Choose units so that the side of the square is length 1.
Then the perimeter equals 4 units and the area equals $1\times 1=1$ unit.
Thus, the perimeter of the square is four times its area.

 
S4. ${\mathbf{0 = 1}}$  

Proof:
$0=0+0+0+\cdots$. But $0=1-1$, so

$$0=(1-1)+(1-1)+(1-1)+\cdots$$ So, by rearranging the brackets, we have $$0=1+(-1+1)+(-1+1)+(-1+1)+\cdots = 1+0+0+0+\cdots = 1$$


 
MAIN QUESTIONS  
 
M1. ${\mathbf{\infty = -1}}$

Proof:
Let $$x=1+2+4+8+\dots$$ Thus, $$1+2x = 1+2(1+2+4+\cdots) = 1+2+4+8+\cdots = x$$ Thus, $1+2x=x$. Rearranging this gives $x=-1$. However, $x$ is also obviously infinite. Thus, $\infty = -1$.

 
M2. Any two real numbers are the same

Proof:
Pick any three real numbers $a$, $b$ and $c$.
If $a^b = a^c$, then $b = c$.
Therefore, since $1^x = 1^y$, we may deduce $x = y$ for any two real numbers $x$ and $y$.


M3. All numbers are equal

Proof:
Suppose that all numbers were not the same. Choose two numbers $a$ and $b$ which are not the same. Therefore one is bigger; we can suppose that $a> b$. Therefore, there is a positive number $c$ such that $a=b+c$. Therefore, multiplying sides by $(a-b)$ gives

$$a(a-b) = (b+c)(a-b)$$ Expanding gives $$a^2-ab = ab-b^2+ac -bc$$ Rearranging gives $$a^2-ab-ac= ab-b^2-bc$$ Taking out a common factor gives $$a(a-b-c) = b(a-b-c)$$ Dividing throughout by $(a-b-c)$ gives $a=b$, therefore $a$ and $b$ could not have been different after all, hence all numbers are equal.

 
 
M4. All numbers are equal - version 2    

Proof:
Choose any two numbers $a$ and $b$, where $a \neq b$, and let $a+b=s$.

Thus, $(a+b)(a-b) = s(a-b)$

Thus, $a^2-b^2 = sa - sb$

Thus, $a^2 -sa = b^2-sb$

Thus, $a^2-sa+s^2/4 = b^2-sb+s^2/4$

Thus, $(a-s/2)^2 = (b-s/2)^2$

Thus, $a-s/2 = b-s/2$

Thus, $a=b$, and so all numbers are equal

PUDDING QUESTIONS  
 

P1. ${\mathbf{3 = 0}}$

Proof:
Consider the quadratic equation $x^2+x+1=0$. Then, we can see that $x^2=-x-1$. Assuming that $x$ is not zero (which it clearly isn't, from the equation) we can divide by $x$ to give

$$x= -1-\frac{1}{x}$$ Substitute this back into the $x$ term in the middle of the original equation, so $$x^2 +\left(-1-\frac{1}{x}\right)+1=0$$ This reduces to $$x^2=\frac{1}{x}$$ So, $x^3=1$, so $x=1$ is the solution. Substituting back into the equation for $x$ gives $$1^2+1+1=0$$ Therefore, $3=0$. 
 
 

 
P2. The smallest positive number is 1 

Proof:
Suppose that $x$ is the smallest positive number. Clearly $x\le 1$ and also $x^2> 0$. Since $x$ is the smallest positive number, $x^2$ can't be smaller then $x$, so we must have $x^2\geq x$. We can divide both sides of this by the positive number $x$ to get $x \geq 1$. Since $x$ is both less than or equal to $1$ and greater than or equal to $1$, $x$ must equal $1$. Thus the smallest positive number is $1$. 
 
 
 
P3. ${\mathbf{1=-1}}$.

Proof:
Clearly, $-1=-1$ and $\frac{1}{1} = \frac{-1}{-1}$
Therefore, $-1\times \frac{1}{1}=-1\times \frac{-1}{-1}$
Therefore, $\frac{-1\times 1 }{1}=\frac{-1\times -1}{-1}$
Therefore, $\frac{-1}{1}=\frac{1}{-1}$
Therefore, $\sqrt{\frac{-1}{1}}=\sqrt{\frac{1}{-1}}$
Therefore, $\frac{\sqrt{-1}}{1}=\frac{1}{\sqrt{-1}}$
Multipliying both sides by $\sqrt{-1}\times 1$ gives $\sqrt{-1}\times \sqrt{-1} = 1\times 1$
Therefore, $-1=1$