Problems

There are $24$ children in the class and some of them are friends with each other. The following rules apply:

• If someone (say Alice) is a friend with someone else (say Bob), then the second student (Bob) is also a friend with the first (Alice).

• If Alice is friend with Bob and Bob is friend with Claire, then Alice is also friend with Claire.

Find a misconception in the following statement: under the above conditions Alice is friend with herself.

Theorem: All people have the same eye color.

"Proof" by induction: This is clearly true for one person.

Now, assume we have a finite set of people, denote them as $a_1,a_2,...,a_n$, and the inductive hypothesis is true for all smaller sets. Then if we leave aside the person $a_1$, everyone else $a_2,a_3,...,a_n$ has the same color of eyes and if we leave aside $a_n$, then all $a_1,a_2,a_3,...,a_{n-1}$ also have the same color of eyes. Thus any $n$ people have the same color of eyes.
Find a mistake in this "proof".

Let’s prove that $1$ is the largest natural number.
Let $n$ be the largest natural number. Then, $n^2$, being a natural number, is less than or equal to $n$. Therefore $n^2-n=n(n-1)\le 0$. Hence, $0\le n\le 1$. Therefore $n=1$.

Find the mistake in the sequence of equalities: $-1=(-1{)}^{\frac{2}{2}}=((-1{)}^2{)}^{\frac{1}{2}}=1^{\frac{1}{2}}=1$.

Theorem: If we mark $n$ points on a circle and connect each point to every other point by a straight line, the lines divide the interior of the circle is into is $2n-1$ regions.
"Proof": First, let’s have a look at the smallest natural numbers.

• When $n=1$ there is one region (the whole disc).

• When $n=2$ there are two regions (two half-discs).

• When $n=3$ there are $4$ regions (three lune-like regions and one triangle in the middle).

• When $n=4$ there are $8$ regions, and if you’re still not convinced then try $n=5$ and you’ll find $16$ regions if you count carefully.

Our proof in general will be by induction on $n$. Assuming the theorem is true for $n$ points, consider a circle with $n+1$ points on it. Connecting $n$ of them together in pairs produces $2n-1$ regions in the disc, and then connecting the remaining point to all the others will divide the previous regions into two parts, thereby giving us $2\times (2n-1)=2n$ regions.

Let’s "prove" that the number $1$ is a multiple of $3$. We will use the symbol $\equiv$ to denote "congruent modulo $3$". Thus, what we need to prove is that $1\equiv 0$ modulo $3$. Let’s see: $1\equiv 4$ modulo $3$ means that $2^1\equiv 2^4$ modulo $3$, thus $2\equiv 16$ modulo $3$, however $16$ gives the remainder $1$ after division by $3$, thus we get $2\equiv 1$ modulo $3$, next $2-1\equiv 1-1$ modulo $3$, and thus $1\equiv 0$ modulo $3$. Which means that $1$ is divisible by $3$.

Recall that $(n+1{)}^2=n^2+2n+1$ and after expansion we get $(n+1{)}^2-(2n+1)=n^2$. Subtract $n(2n+1)$ from both sides $(n+1{)}^2-(2n+1)-n(2n+1)=n^2-n(2n+1)$ and rewrite it as $(n+1{)}^2-(n+1)(2n+1)=n^2-n(2n+1)$.
Now we add $\frac{(2n+1{)}^2}{4}$ to both sides: $(n+1{)}^2-(n+1)(2n+1)+\frac{(2n+1{)}^2}{4}=n^2-n(2n+1)+\frac{(2n+1{)}^2}{4}$.
Factor both sides into square: $((n+1)-\frac{2n+1}{2}{)}^2=(n-\frac{2n+1}{2}{)}^2$.
Now take the square root: $(n+1)-\frac{2n+1}{2}=n-\frac{2n+1}{2}$.
Add $\frac{2n+1}{2}$ to both sides and we get $n+1=n$ which is equivalent to $1=0$.

Look at the following diagram, depicting how to get an extra cell by reshaping triangle.

Can you find a mistake? Certainly the triangles have different area, so we cannot obtain one from the other one by reshaping.

This problem is often called "The infinite chocolate bar". Depicted below is a way to get one more piece of chocolate from the $5\times 6$ chocolate bar. Do you see where is it wrong?

Consider the following "proof" that any triangle is equilateral: Given a triangle $ABC$, we first prove that $AB=AC$. First let’s draw the bisector of the angle $\mathrm{\angle }A$. Now draw the perpendicular bisector of segment $BC$, denote by $D$ the middle of $BC$ and by $O$ the intersection of these lines. See the diagram

Draw the lines $OR$ perpendicular to $AB$ and $OQ$ perpendicular to $AC$. Draw lines $OB$ and $OC$. Then the triangles, $RAO$ and $QAO$ are equal, since we have equal angles $\mathrm{\angle }ORA=\mathrm{\angle }OQA=90°,$ and $\mathrm{\angle }RAO=\mathrm{\angle }QAO,$ and the common side $AO$. On the other hand the triangles $ROB$ and $QOC$ are also equal since the angles $\mathrm{\angle }BRO=\mathrm{\angle }CQO=90°$, the hypotenuses $BO=OC$ the legs $RO=OQ$. Thus, $AR=AQ,$ $RB=QC,$ and $AB=AR+RB=AQ+QC=AC.$ Q.E.D.

As a corollary, one can show that all the triangles are equilateral, by showing that $AB=BC$ in the same way.