Chapter 3
A Gentle Introduction to the Art of Mathematics · 32 exercises
Problem 1
Prove that if the cube of an integer is odd, then that integer is odd.
6 step solution
Problem 2
Prove that every prime number other than 2 and 3 has the form \(6 q+1\) or \(6 q+5\) for some integer \(q\). (Hint: this problem involves thinking about cases as well as contrapositives.)
5 step solution
Problem 2
Recall that a quadratic equation \(a x^{2}+b x+c=0\) has two real solutions if and only if the discriminant \(b^{2}-4 a c\) is positive. Prove that if \(a\) and \(c\) have different signs then the quadratic equation has two real solutions.
5 step solution
Problem 2
Prove that 129 is odd.
4 step solution
Problem 2
Prove that whenever a prime \(p\) does not divide the square of an integer, it also doesn't divide the original integer. \(\left(p \nmid x^{2} \Longrightarrow p \nmid x\right)\)
4 step solution
Problem 3
Show that the sum of any three consecutive integers is divisible by 3 .
5 step solution
Problem 3
The alternating sum of factorials provides an interesting example of a sequence of integers. $$ \begin{array}{c} 1 !=1 \\ 2 !-1 !=1 \\ 3 !-2 !+1 !=5 \\ 4 !-3 !+2 !-1 !=19 \end{array} $$ et cetera Are they all prime? (After the first two 1 's.)
5 step solution
Problem 3
Prove (by contradiction) that there is no largest integer.
5 step solution
Problem 3
Prove that the sum of two rational numbers is a rational number
4 step solution
Problem 4
There is a graph known as \(K_{4}\) that has 4 nodes and there is an edge between every pair of nodes. The pebbling number of \(K_{4}\) has to be at least 4 since it would be possible to put one pebble on each of 3 nodes and not be able to reach the remaining node using pebbling moves. Show that the pebbling number of \(K_{4}\) is actually \(4 .\)
4 step solution
Problem 4
Prove that for all integers \(a, b,\) and \(c,\) if \(a \mid b\) and \(a \mid(b+c),\) then \(a \mid c\).
4 step solution
Problem 4
Prove that the sum of an odd number and an even number is odd.
6 step solution
Problem 4
Prove (by contradiction) that there is no smallest positive real number.
4 step solution
Problem 5
Prove that if the sum of two integers is even, then so is their difference.
6 step solution
Problem 5
Show that if \(x\) is a positive real number, then \(x+\frac{1}{x} \geq 2\).
5 step solution
Problem 5
Prove (by contradiction) that the sum of a rational and an irrational number is irrational.
6 step solution
Problem 6
A vampire number is a \(2 n\) digit number \(v\) that factors as \(v=x y\) where \(x\) and \(y\) are \(n\) digit numbers and the digits of \(v\) are precisely the digits in \(x\) and \(y\) in some order. The numbers \(x\) and \(y\) are known as the "fangs" of \(v\). To eliminate trivial cases, both fangs can't end with \(0 .\) Show that there are no 2 -digit vampire numbers. Show that there are seven 4 -digit vampire numbers.
6 step solution
Problem 6
Prove that for all real numbers \(a, b,\) and \(c,\) if \(a c<0,\) then the quadratic equation \(a x^{2}+b x+c=0\) has two real solutions.
5 step solution
Problem 6
Prove (by contraposition) that for all integers \(x\) and \(y,\) if \(x+y\) is odd, then \(x \neq y\).
6 step solution
Problem 7
Lagrange's theorem on representation of integers as sums of squares says that every positive integer can be expressed as the sum of at most 4 squares. For example, \(79=7^{2}+5^{2}+2^{2}+1^{2} .\) Show (exhaustively) that 15 can not be represented using fewer than 4 squares.
6 step solution
Problem 7
Show that \(\left(\begin{array}{c}n \\ k\end{array}\right) \cdot\left(\begin{array}{c}k \\ r\end{array}\right)=\left(\begin{array}{c}n \\\ r\end{array}\right) \cdot\left(\begin{array}{c}n-r \\\ k-r\end{array}\right)\) (for all integers \(r, k\) and \(n\) with \(r \leq k \leq n)\).
6 step solution
Problem 7
Prove that if \(x\) is an odd integer, then \(x^{2}\) is of the form \(4 k+1\) for some integer \(k\).
4 step solution
Problem 7
Prove (by contraposition) that for all real numbers \(a\) and \(b\), if \(a b\) is irrational, then \(a\) is irrational or \(b\) is irrational.
6 step solution
Problem 8
In proving the product rule in Calculus using the definition of the derivative, we might start our proof with: $$\begin{array}{c} \frac{\mathrm{d}}{\mathrm{d} x}(f(x) \cdot g(x)) \\ =\lim _{h \longrightarrow 0} \frac{f(x+h) \cdot g(x+h)-f(x) \cdot g(x)}{h} \end{array}$$ The last two lines of our proof should be: $$\begin{array}{c} =\lim _{h \longrightarrow 0} \frac{f(x+h)-f(x)}{h} \cdot g(x)+f(x) \cdot \lim _{h \longrightarrow 0} \frac{g(x+h)-g(x)}{h} \\ =\frac{\mathrm{d}}{\mathrm{d} x}(f(x)) \cdot g(x)+f(x) \cdot \frac{\mathrm{d}}{\mathrm{d} x}(g(x)) \end{array}$$ Fill in the rest of the proof.
5 step solution
Problem 8
A Pythagorean triple is a set of three natural numbers, \(a, b\) and \(c,\) such that \(a^{2}+b^{2}=c^{2}\). Prove that, in a Pythagorean triple, at least one of \(a\) and \(b\) is even. Use either a proof by contradiction or a proof by contraposition.
8 step solution
Problem 9
The trichotomy property of the real numbers simply states that every real number is either positive or negative or zero. Trichotomy can be used to prove many statements by looking at the three cases that it guarantees. Develop a proof (by cases) that the square of any real number is non-negative.
6 step solution
Problem 9
True or false: Whenever an integer \(n\) is a divisor of the square of an integer, \(m^{2},\) it follows that \(n\) is a divisor of \(m\) as well. (In symbols, \(\left.\forall n \in \mathbb{Z}, \forall m \in \mathbb{Z}, n\left|m^{2} \Longrightarrow n\right| m .\right)\) Prove your answer.
5 step solution
Problem 9
Prove that \(\forall x \in \mathbb{R}, x \notin \mathbb{Z} \Longrightarrow\lfloor x\rfloor+\lfloor-x\rfloor=-1\).
5 step solution
Problem 10
In an exercise in Section 3.2 we proved that the quadratic equation \(a x^{2}+b x+c=0\) has two solutions if \(a c<0 .\) Find a counterexample which shows that this implication cannot be replaced with a biconditional.
6 step solution
Problem 11
Suppose that \(a, b\) and \(c\) are integers such that \(a \mid b\) and \(b \mid c .\) Prove that \(a \mid c\)
4 step solution
Problem 12
Suppose that \(a, b, c\) and \(d\) are integers with \(a \neq c .\) Further, suppose that \(x\) is a real number satisfying the equation $$ \frac{a x+b}{c x+d}=1 $$ Show that \(x\) is rational. Where is the hypothesis \(a \neq c\) used?
7 step solution
Problem 13
Show that if two positive integers \(a\) and \(b\) satisfy \(a \mid b\) and \(b \mid a\) then they are equal.
6 step solution