Prove that when the square of a positive odd integer is divided by 4 the remainder is always 1.
$(2n+1)^2 = 4(n^2+n)+1$
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.
- $\Leftarrow$
- $\Rightarrow$
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.
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.
Disprove the statement The sum of two consecutive prime numbers is always even.
by use of a counterexample. $2+3$
Use proof by contradiction to prove that there are an infinite number of prime numbers.
Assume there is a finite number of prime numbers. <br> Add $1$ to the product of all the prime numbers <br> This new number does not have any of the primes we know as factors <br> It is either prime, or there is a prime factor of it that we do not know about <br> In both cases we have found a new prime number
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 <br> $\frac{a^2}{b^2} = 2 \Leftrightarrow a^2 = 2b^2$ <br> $a^2$, and $a$, are both even. <br> Write $a = 2n \Rightarrow 2b^2 = 4n^2 \Leftrightarrow b^2=2n^2$ <br>> $b$ is even <br> This is a contradiction because $a$ and $b$ have no common factors, so $\sqrt{2}$ cannot be rational
Prove by contradiction that for all real $\theta$, $$\cos\theta + \sin\theta \leq \sqrt{2}$$
Assume $\cos\theta + \sin\theta > \sqrt{2}$ <br> $\Rightarrow 1 + \sin2\theta > 2$ <br> $\Rightarrow \sin2\theta> 1$ <br> This is a contradiction so the assumption cannot be true.
Prove by contradiction that for all real $x$ $$(13x+1)^2 + 3 > (5x-1)^2$$
Assume $(13x+1)^2 + 3 \leqslant (5x-1)^2$ <br> $\Rightarrow 144x^2+36x+3 \leqslant 0$ <br> $\Rightarrow 48(x+0.125)^2+0.25 \leqslant 0$ <br> Contradiction because both terms are positive.
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: <br> $(3k+1)^2 - 1 = 3(3k^2+2k)$ <br> $(3k-1)^2=3(3k^2-2k)$ and every case is divisible by 3
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 $4$ <br> $a+b = 4m$ and $a-b = 4n$ <br> $2a = 4m+4n$ and $a = 2(m+n)$ <br> $a$ is not odd and this is a contraction.
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. <br> If not, then $\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}} = \sqrt{2}^2 = 2$ which is rational
- 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 $\sqrt{4pq}$
- $p = q = -1$
Prove by contradiction that $\log_{10}5$ is irrational.
Assume it is rational <br> $\log 5 = \frac{a}{b}$ where $b \neq 0$ <br> $10^a = 5^b \Rightarrow 2^a5^a=5^b$ <br> This is only true if $a = 0 = b$ so this is a contraction.
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. <br> $(3n-1)^2 = 3(3n^2-2n)+1$ <br> $(3n)^2 = 3(3n^2)$ <br> $(3n+1)^2 = 3(3n^2+2n)+1$
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 $b=2q+1$ <br> $a^2+b^2 = 2(2p^2+2p+2q^2+2q+1)$ <br> This means $c^2$ and $c$ are even <br> If $c = 2r$, $c^2 = 4r^2$ which is a multiple of $4$ <br> $a^2+b^2$ is only a multiple of $2$, so contradiction
Show that the sum of cubes of any three consecutive positive integers is a multiple of $9$.
Let $n=1$ be the first number. <br> Sum: $3n^3+6n = 3n(n^2+2)$ <br> $n = 3k-1$ gives $9(3k-1)(3k^2-2k+1)$ <br> $n = 3k$ gives $9k(9k^2+2)$ <br> $n = 3k+1$ gives $9(3k+1)(3k^2+2k+1)$ <br> Every case is a multiple of 9.
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. <br> Result: $n^4+6n^3+11n^2+6n+1$ <br> This is equal to $(n^2+kn+1)^2 \Rightarrow k = 3$