MathsQuestions

A Level
Proof

1. Prove that when the square of a positive odd integer is divided by 4 the remainder is always 1.

$(2n+1)^2 = 4(n^2+n)+1$

2. Use $\Leftarrow$, $\Rightarrow$ or $\Leftrightarrow$ to link the following statements.

$S$ is a rectangle. $S$ is a square.
$n$ is even. $n$ is an integer.

$\Leftarrow$
$\Rightarrow$

3. A student writes $$x^2-2x=0 \Leftrightarrow x-2=0 \Leftrightarrow x = 2$$ Determine if the student has made a mistake.

The first $\Leftrightarrow$ should be $\Rightarrow$, so the student has made a mistake.

4. Disprove the following by a counterexample: $$|2x + 1| \leq 5 \Rightarrow |x| \leq 2$$

When $x = -3$ the first inequality is true but the second is not.

5. Disprove the statement

The sum of two consecutive prime numbers is always even.

by use of a counterexample.

$2+3$

6. Use proof by contradiction to prove that there are an infinite number of prime numbers.

Assume there is a finite number of prime numbers.
Add $1$ to the product of all the prime numbers
This new number does not have any of the primes we know as factors
It is either prime, or there is a prime factor of it that we do not know about
In both cases we have found a new prime number

7. Prove by contradiction that $\sqrt{2}$ is irrational.

Assume $\sqrt{2}$ is rational and $\sqrt{2} = \frac{a}{b}$ where $a$ and $b$ have no common factors
$\frac{a^2}{b^2} = 2 \Leftrightarrow a^2 = 2b^2$
$a^2$, and $a$, are both even.
Write $a = 2n \Rightarrow 2b^2 = 4n^2 \Leftrightarrow b^2=2n^2$
$b$ is even
This is a contradiction because $a$ and $b$ have no common factors, so $\sqrt{2}$ cannot be rational

8. Prove by contradiction that for all real $\theta$, $$\cos\theta + \sin\theta \leq \sqrt{2}$$

Assume $\cos\theta + \sin\theta > \sqrt{2}$
$\Rightarrow 1 + \sin2\theta > 2$
$\Rightarrow \sin2\theta> 1$
This is a contradiction so the assumption cannot be true.

9. Prove by contradiction that for all real $x$ $$(13x+1)^2 + 3 > (5x-1)^2$$

Assume $(13x+1)^2 + 3 \leqslant (5x-1)^2$
$\Rightarrow 144x^2+36x+3 \leqslant 0$
$\Rightarrow 48(x+0.125)^2+0.25 \leqslant 0$
Contradiction because both terms are positive.

10. Prove by exhaustion that if $n$ is a positive integer that is not divisible by 3, then $n^2-1$ is divisible by 3.

Two cases if not divisible by 3:
$(3k+1)^2 - 1 = 3(3k^2+2k)$
$(3k-1)^2=3(3k^2-2k)$ and every case is divisible by 3

11. Prove by contradiction that if $a$ and $b$ are positive odd integers, where $a > b$ and $a + b$ is a multiple of $4$, then $a - b$ cannot be a multiple of $4$.

Assume both $a+b$ and $a-b$ are multiples of $4$
$a+b = 4m$ and $a-b = 4n$
$2a = 4m+4n$ and $a = 2(m+n)$
$a$ is not odd and this is a contraction.

12. By considering $\sqrt{2}^{\sqrt{2}}$, prove that an irrational number raised to the power of an irrational number can be a rational number.

If it is rational, then it has been proved.
If not, then $\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}} = \sqrt{2}^2 = 2$ which is rational

13.

Prove that $p+q\geqslant\sqrt{4pq}$ if $p$ and $q$ are both positive.
Use a counterexample to prove that this is not true if both are negative.

LHS: $\sqrt{(p-q)^2+4pq}$, which is always greater than $\sqrt{4pq}$
$p = q = -1$

14. Prove by contradiction that $\log_{10}5$ is irrational.

Assume it is rational
$\log 5 = \frac{a}{b}$ where $b \neq 0$
$10^a = 5^b \Rightarrow 2^a5^a=5^b$
This is only true if $a = 0 = b$ so this is a contraction.

15. Prove that the square of a positive integer can never be of the form $3k + 2$, where $k$ is an integer.

Positive integer is 1 less than 3, a multiple of 3, or 1 more than 3.
$(3n-1)^2 = 3(3n^2-2n)+1$
$(3n)^2 = 3(3n^2)$
$(3n+1)^2 = 3(3n^2+2n)+1$

16. Given $a^2 + b^2 = c^2$, where $a$, $b$, and $c$, are all integers, show that $a$ and $b$ cannot both be odd.

Assume both are odd: $a = 2p+1$ and $b=2q+1$
$a^2+b^2 = 2(2p^2+2p+2q^2+2q+1)$
This means $c^2$ and $c$ are even
If $c = 2r$, $c^2 = 4r^2$ which is a multiple of $4$
$a^2+b^2$ is only a multiple of $2$, so contradiction

17. Show that the sum of cubes of any three consecutive positive integers is a multiple of $9$.

Let $n=1$ be the first number.
Sum: $3n^3+6n = 3n(n^2+2)$
$n = 3k-1$ gives $9(3k-1)(3k^2-2k+1)$
$n = 3k$ gives $9k(9k^2+2)$
$n = 3k+1$ gives $9(3k+1)(3k^2+2k+1)$
Every case is a multiple of 9.

18. Prove that if $1$ is added to the product of any four consecutive integers, the result is always a square number.

Let $n$ be the first number.
Result: $n^4+6n^3+11n^2+6n+1$
This is equal to $(n^2+kn+1)^2 \Rightarrow k = 3$