Dodgy proofs

6384 /www/nrich/html/content/id/6384/ 1 2011-07-19T08:40:06 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">Consider these dodgy proofs. Although the results are obviously wrong, where, precisely, do the 'proofs' break down? There are five fun introductory questions, four main questions and four extension questions. Good luck! INTRODUCTORY QUESTIONS F1. A pound equals a penny Proof: £$1 = 100 p = (10p)^2 = ($£$ 0.1)^2 = $£$0.01 = 1p$ F2. ${\mathbf{2 = 3}}$ Proof: $2\times 0 =0$, so $2=0\div 0$. Also, $3\times 0 = 0$, so $3=0 \div 0$. Since $0\div 0$, however it is defined, is clearly the same as $0\div 0$, we must have $2=3$. F3. 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. F4. ${\mathbf{0 = 1}}$ 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 $$ 0=1+(-1+1)+(-1+1)+(-1+1)+\cdots = 1+0+0+0+\cdots = 1 $$ F5. There is a living man with at least four heads Proof: No man has three heads. Any living man has at least one head more than no man. Therefore, there is a living man with at least four heads. MAIN QUESTIONS M1. 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 $y$. M2. ${\mathbf{\infty = -1}}$ 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, $\infty = -1$. M3. All numbers are the same 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 $a&gt; b$. 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 through gives $a=b$, therefore $a$ and $b$ could not have been different after all, hence all numbers are the same. M4. All numbers are equal Proof: Choose any two numbers $a$ and $b$ 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, $a=b$ E1. ${\mathbf{3 = 0}}$ 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, $3=0$. E2. The smallest positive number is 1 Proof: Suppose that $x$ is the smallest positive number. Clearly $x\le 1$ and also $x^2&gt; 0$. 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 $1$. E3. ${\mathbf{1=-1}}$. 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, $-1=1$ E4. All cows in a field are the same colour Proof : by induction on the number of cows. Induction hypothesis: n cows in a field are the same colour, for all n=1,2,3,4,... Initial Step: Clearly one cow in a field is the same colour as itself, so the induction hypothesis is true for n=1. Inductive Step: Now suppose n is at least 1, we have n cows in a field F, and that the induction hypothesis has been proved for all fields containing at most n cows. By the induction hypothesis all the cows in F are the same colour. Now, remove any cow from F and put it to one side. Then take a cow from some other field and put it in field F. We again have n cows in field F, so by the induction hypothesis they are all the same colour. Finally, put back the first cow you thought of. We already know that it's the same colour as all the other cows in F, and so now we have n+1 cows in field F, and they're all the same colour. This completes the induction step! </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do this problem?</h3> Proof is, of course, a central part of mathematics. However constructing proofs is often difficult for novices. This problem provides a bridge through the device of classic faulty proofs of 'obviously' wrong results. Analysing these faulty proofs will provide training in reading proofs, raise awareness of mathematical hazards, such as division by zero, and provide motivation for rigour. You might wish to use some of these proof every so often throughout the year. <h3>Possible approach</h3> <div>Make it a race to find errors in as many proofs as possible in the available time. Split the class into small groups and give them one of the first three proofs to work on. Let everyone know that the next proof will be given out when the group has convinced you of the location of the error.</div> <div> </div> <div>When a group wishes to try to convince you of an error choose one member of the group at random: only this member is permitted to speak or write. If you are convinced, give the group the next dodgy proof; otherwise give minimal feedback such as "Sorry, I'm not convinced" or "That explanation didn't seem clear to me" and leave them to attempt to tighten up their argument.</div> <div> </div> <div>Use common sense to judge the acceptable level of rigour.</div> <div> </div> <div>Be on the lookout for the tell-tale signs of 'um', 'er', 'the thing' and 'it' which so often indicate confused thinking.</div> <div> </div> <div> </div> <div> </div> <h3>Key questions</h3> <div>Does that argument sound clear to you? (students will probably know when they are bluffing!)</div> <div> </div> <h3>Possible extension</h3> <div>Students might be asked to invent their own dodgy proofs.</div> <div> </div> <h3>Possible support</h3> <div>You could simply discuss the proofs as a group or you could draft in some helpers who assist you in deciding if explanations are acceptably clear.</div> </mdoxml> 2 5 0 0 0 0 1 Dodgy proofs These proofs are wrong. Can you see why? Using, Applying and Reasoning about Mathematics Mathematical reasoning & proof