Problems
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 \(\angle A\). 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 \(\angle ORA = \angle OQA = 90°,\) and \(\angle RAO = \angle QAO,\) and the common side \(AO\). On the other hand the triangles \(ROB\) and \(QOC\) are also equal since the angles \(\angle BRO = \angle CQO = 90°\), 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.
Let’s prove the following statement: every graph without isolated vertices is connected.
Proof We use the induction on the number of vertices. Clearly the statement is true for graphs with \(2\) vertices. Now, assume we have proven the statement for graphs with up to \(n\) vertices.
Take a graph with \(n\) vertices by induction hypothesis it must be connected. Let’s add a non-isolated vertex to it. As this vertex is not isolated, it is connected to one of the other \(n\) vertices. But then the whole graph of \(n+1\) vertices is connected!
Each number denotes the area of a rectangle it is written into. What is the area of the last rectangle?
Divide the trapezium into two parts such that they can be reassembled to make a triangle