A number \(n\) is an integer such that \(n\) is not divisible by \(3\) or by \(2\). Show that \(n^2-1\) is divisible by \(24\).
Show that for any two positive real numbers \(x,y\) it is true that \(x^2+y^2 \ge 2xy\).
Find all pairs of integers \((x,y)\) so that the following equation is true \(xy = y+x\).
The perimeter of the triangle \(\triangle ABC\) is \(10\). Let \(D,E,F\) be the midpoints of the segments \(AB,BC,AC\) respectively. What is the perimeter of \(\triangle DEF\)?
Show that if \(1+2+\dots+n = \frac{n(n+1)}{2}\), then \(1+2+\dots+(n+1) = \frac{(n+1)((n+1)+1)}{2}\).
Show that \(1+2+\dots+n = \frac{n(n+1)}{2}\) for every natural number \(n\).
Show that if \(1+2^1+2^2+\dots+2^{10} = 2^{11} - 1\), then \(1+2^1+2^2+\dots+2^{11} = 2^{12} - 1\).
Show that \(1+2^1+2^2+\dots+2^n = 2^{n+1} - 1\) for every natural number \(n\).
What is wrong with the following proof that “all rulers have the same length" using induction?
Base case: suppose that we have one ruler, then clearly it clearly has the same length as itself.
Assume that any \(n\) rulers have the same length for the induction hypothesis. If we have \(n+1\) rulers, the first \(n\) ruler have the same length by the induction hypothesis, and the last \(n\) rulers have the same length also by induction hypothesis. The last ruler has the same length as the middle \(n-1\) rulers, so it also has the same length as the first ruler. This means all \(n+1\) rulers have the same length.
By the principle of mathematical induction, all rulers have the same length.