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Again, by induction, all remaining humans are equal

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EXERCISES 6: LECTURE FOUNDATIONS OF MATHEMATICS

Exercise 1. Explain in detail what the mistake in the following argument is.

Claim: Assume thatn≥1 humans are in a room. Then allnhumans are equal.

Proof: By induction. First, ifn = 1, then the claim is clearly true. (Everyone is equal to him/herself.) So assume that the claim is true for somen, and assume that n+ 1 are in a room. If one human leaves the room, then all the remaining humans are equal, by induction.

So let the one human reenter the room, and let another human leave the room. Again, by induction, all remaining humans are equal. Hence, alln+ 1humans are equal. ?

Exercise 2. LetX be a set containing nelements. Show that its power setP(X) contains 2n elements.

Exercise 3. Prove the following statements by induction, wheren∈N0. (a) Pn

k=0k= n(n+1)2 . (b) Pn

k=0k2= n(n+1)(2n+1)

6 .

(c) Forn≥2one has n+ 1<2n. (d) n3−nis divisible by 3.

(e) nk−nis divisible byk∈N. (Errata: Beware, this exercise was flawed. It is only true for kbeing a prime number, but e.g. fork= 4it is possible to construct counterexamples.)

Exercise 4. Letp∈N0, p >1. Show thatp is prime if and only if (p|ab)⇒(p|a∨p|b) ∀a, b∈N0.

Submission of the exercise sheet: 05.Nov.2018 before the lecture. Return of the exercise sheet: 08.Nov.2018 during the exercise sessions.

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Again, by induction, all remaining humans are equal

If one human leaves the room, then all the remaining humans are equal, by induction.. So let the one human reenter the room, and let another human leave