What is common between the two examples above? In fact, if you want to know some fancy words (you should understand what they mean, of course), we just stated that a direct proof and a proof by contrapositive is the same thing. In simple words it means that “If A then B” is the same thing as “If not B, then not A”.
A proof by contrapositive can be very useful. In some problems it is much easier to prove “If not B, then not A” compare to “If A then B”. Let’s consider another example, where a proof by contrapositive can be very useful
There are 10 lines drawn on the plane, all intersecting at the same point. Show that there will be at least two lines with angle between them less than \(18^o\).
Is “If you are not mad, then you growl when you are angry and wag your tail when you are pleased” the same thing as “If you don’t growl when you are angry or don’t wag your tail when you are pleased, then you are mad”?
The cat and Alice ate three cakes. Show that one of them ate at least two cakes.
You are given \(11\) natural numbers. Show that you can choose two among those numbers such that the difference between the chosen numbers is divisible by \(10\).
A rectangle \(5 \times 9\) is cut into 10 small rectangles with sides of integer lengths. Show that there are two identical rectangles among them.
Let \(n!= n\times (n-1) \times(n-2)\times \dots \times 2\times 1\). Prove that if \(n!+1\) is divisible by \(n+1\), then \(n+1\) is a prime number.
Denote by \(\overline{ab} = 10a +b\) the two-digit number whose first and second digits are \(a\) and \(b\) respectively. Do there exist two \(2\)-digit numbers \(\overline{ab}\) and \(\overline{cd}\) such that \(\overline{ab} \times \overline{cd} = \overline{abcd}\)? (Here \(\overline{abcd}\) is a four-digit number with digits \(a\), \(b\), \(c\) and \(d\), i.e. \(\overline{abcd} = 1000a + 100b +10c +d\).)
Sixty children came to a maths circle at UCL. Among any ten children who came to the circle there are three from the same school. Show that there are 15 children from the same school among all the children who came to the maths circle.
The people in Wonderland are having an election. Every voter writes 10 candidate names on a bulletin and puts it in a ballot box.
There are 11 ballot boxes all together. The March Hare, who is counting the votes, is very surprised to discover that there is at least one bulletin in each ballot box. Moreover, he learned that if he takes one bulletin from each ballot box (11 bulletins all together), then there is always a candidate whose name is written in each of the 11 chosen bulletins. Prove that there is a ballot box, in which all the bulletins contain the name of the same candidate.