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# Dodgy Proofs

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Age 16 to 18

Challenge Level

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, $-1=1$

A and B are two fixed points on a circle and RS is a variable diamater. What is the locus of the intersection P of AR and BS?

Make a conjecture about the sum of the squares of the odd positive integers. Can you prove it?