Language and Illustrations of Proofs

Proof, EXT2 P1 2025 SPEC2 1 MC

A tiger is a type of cat.

Consider the following statement.

'If I have a tiger, then I have a cat.'

The contrapositive of this statement is

if I do not have a tiger, then I do not have a cat.
if I have a cat, then I have a tiger.
if I do not have a cat, then I do not have a tiger.
if I do not have a tiger, then I have a different type of cat.

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\(C\)

Show Worked Solution

\(\text{Statement: If} \ \ X \Rightarrow Y\)

\(\text{Contrapositive: If}\ \ \neg \ Y\ \Rightarrow \neg \ X\)

\(\text{If I don’t have a cat} \ \ \Rightarrow \ \ \text{I don’t have a tiger.}\)

\(\Rightarrow C\)

Proof, EXT2 P1 2025 HSC 16a

Consider the equation

\(z^n \cos\left[n \theta\right]+z^{n-1} \cos \left[(n-1) \theta\right]+z^{n-2} \cos \left[(n-2) \theta\right]+\cdots+z\, \cos\left[\theta\right]=1\)

where \(z \in \mathbb{C} , \theta \in \mathbb{R} \), and \(n\) is a positive integer.

Using a proof by contradiction and the triangle inequality, or otherwise, prove that all the solutions to the equation lie outside the circle \(\abs{z}=\dfrac{1}{2}\) on the complex plane. (4 marks)

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\(\text{Proof by contradiction}\)

\(\text{Assume}\ \exists\ z \in \mathbb{C},\ \text{where}\ \abs{z} \in\left[0, \dfrac{1}{2}\right],\ \text{and}\)

\(z^n \cos \left[n \theta \right]+z^{n-1} \cos \left[(n-1) \theta \right] + \ldots +z\, \cos \theta=1\)

\(\text{Using the triangle inequality}\ \ \left(\abs{x}+\abs{y} \geqslant \abs{x+y}\right):\)

\(\left|z^n \cos \left[n \theta\right] \right|+\left|z^{n-1} \cos \left[(n-1) \theta\right] \right|+\ldots+|z\, \cos \theta|\)

\(\geqslant\left|z^n \cos \left[n \theta \right] +z^{n-1} \cos \left[(n-1) \theta \right]+\ldots+z\, \cos \theta\right|\)

\(1 \leqslant\left|z^n \cos \left[n \theta \right]\right|+\left|z^{n-1} \cos \left[(n-1) \theta \right]\right|+\ldots+|z\, \cos \theta|\)

\(1 \leqslant|z|^n+|z|^{n-1}+\cdots+|z| \quad (\text{since}-1 \leqslant \cos (k \theta) \leqslant 1)\)

\(1 \leqslant (\frac{1}{2})^n+(\frac{1}{2})^{n-1}+\cdots+(\frac{1}{2}) \)

\(1 \leqslant \underbrace{2^{-n}+2^{-n+1}+\cdots+2^{-1}}_{\text{GP:}\ a=2^{-n}, r=2}\)

\(1 \leqslant \dfrac{2^{-n}\left(2^n-1\right)}{2-1}\)

\(1 \leqslant 1-2^{-n}\)

\(2^{-n} \leqslant 0 \ \ \text {(which is not true)}\)

\(\therefore \ \text{By contradiction, the original statement is correct}\)

Show Worked Solution

\(\text{Proof by contradiction}\)

\(\text{Assume}\ \exists\ z \in \mathbb{C},\ \text{where}\ \abs{z} \in\left[0, \dfrac{1}{2}\right],\ \text{and}\)

\(z^n \cos \left[n \theta \right]+z^{n-1} \cos \left[(n-1) \theta \right] + \ldots +z\, \cos \theta=1\)

♦♦♦ Mean mark 10%.

\(\text{Using the triangle inequality}\ \ \left(\abs{x}+\abs{y} \geqslant \abs{x+y}\right):\)

\(\left|z^n \cos \left[n \theta\right] \right|+\left|z^{n-1} \cos \left[(n-1) \theta\right] \right|+\ldots+|z\, \cos \theta|\)

\(\geqslant\left|z^n \cos \left[n \theta \right] +z^{n-1} \cos \left[(n-1) \theta \right]+\ldots+z\, \cos \theta\right|\)

\(1 \leqslant\left|z^n \cos \left[n \theta \right]\right|+\left|z^{n-1} \cos \left[(n-1) \theta \right]\right|+\ldots+|z\, \cos \theta|\)

\(1 \leqslant|z|^n+|z|^{n-1}+\cdots+|z| \quad (\text{since}-1 \leqslant \cos (k \theta) \leqslant 1)\)

\(1 \leqslant (\frac{1}{2})^n+(\frac{1}{2})^{n-1}+\cdots+(\frac{1}{2}) \)

\(1 \leqslant \underbrace{2^{-n}+2^{-n+1}+\cdots+2^{-1}}_{\text{GP:}\ a=2^{-n}, r=2}\)

\(1 \leqslant \dfrac{2^{-n}\left(2^n-1\right)}{2-1}\)

\(1 \leqslant 1-2^{-n}\)

\(2^{-n} \leqslant 0 \ \ \text {(which is not true)}\)

\(\therefore \ \text{By contradiction, the original statement is correct}\)

Proof, EXT2 P1 2025 HSC 5 MC

Consider the statement:

If \(x^2-2 x \geq 0\), then \(x \leq 0\).

Which of the following is the contrapositive of the statement?

If \(x>0\), then \(x^2-2 x<0\).
If \(x \leq 0\), then \(x^2-2 x \geq 0\).
If \(x^2-2 x<0\), then \(x<0\).
If \(x^2-2 x \leq 0\), then \(x>0\).

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\(A\)

Show Worked Solution

\(\text{Statement: If}\ \ p\ \Rightarrow \ q\)

\(\text{Contrapositive statement: If}\ \ \neg\ q\ \Rightarrow \ \neg\ p\)

\(\therefore\ \text{If}\ \ x>0\ \ \Rightarrow\ \ x^2-2 x<0\).

\(\Rightarrow A\)

Proof, EXT2 P1 2025 HSC 11e

Prove by contradiction that \(\sqrt{3}+\sqrt{5}>\sqrt{11}\). (2 marks)

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\(\text{Prove} \ \ \sqrt{3}+\sqrt{5}>\sqrt{11}\)

\(\text{Assume}\ \ \sqrt{3}+\sqrt{5}\)	\(\leqslant\sqrt{11}\)
\((\sqrt{3}+\sqrt{5})^2\)	\(\leqslant 11\)
\(3+2 \sqrt{15}+5\)	\(\leqslant 11\)
\(2 \sqrt{15}\)	\(\leqslant 3\)
\(60\)	\(\leqslant 9 \ \text{(incorrect)}\)

\(\therefore \ \text{By contradiction} \ \ \sqrt{3}+\sqrt{5}>\sqrt{11}\)

Show Worked Solution

\(\text{Prove} \ \ \sqrt{3}+\sqrt{5}>\sqrt{11}\)

\(\text{Assume}\ \ \sqrt{3}+\sqrt{5}\)	\(\leqslant\sqrt{11}\)
\((\sqrt{3}+\sqrt{5})^2\)	\(\leqslant 11\)
\(3+2 \sqrt{15}+5\)	\(\leqslant 11\)
\(2 \sqrt{15}\)	\(\leqslant 3\)
\(60\)	\(\leqslant 9 \ \text{(incorrect)}\)

\(\therefore \ \text{By contradiction} \ \ \sqrt{3}+\sqrt{5}>\sqrt{11}\)

Proof, EXT2 P1 2025 HSC 2 MC

Consider the statement:

\(\exists\, x \in Z\), such that \(x^2\) is odd.

Which of the following is the negation of the statement?

\(\forall\, x \in Z , x^2\) is odd
\(\forall\, x \in Z , x^2\) is even
\(x^2\) is even \(\Rightarrow x \in Z\)
\(\exists\, x \in Z\), such that \(x^2\) is even

Show Answers Only

\(B\)

Show Worked Solution

\(x^2\ \text{is odd is negated by}\ x^2\ \text{is even.}\)

\(\therefore\ \text{Negation of statement:}\ \forall\, x \in Z , x^2\ \text{is even}\)

\(\Rightarrow B\)

Proof, EXT2 P1 2024 SPEC2 1 MC

Consider the statement

'for any integers \(m\) and \(n\), if \(m+n \geq 9\) then \(m \geq 5\) or \(n \geq 5\) '.

The contrapositive of this statement is

if \(m<5\) or \(n<5\), then \(m+n<9\)
if \(m \geq 5\) or \(n \geq 5\), then \(m+n \geq 9\)
if \(m<5\) and \(n<5\), then \(m+n<9\)
if \(m \leq 5\) and \(n \leq 5\), then \(m+n \leq 9\)

Show Answers Only

\(C\)

Show Worked Solution

\(\text{Statement: If}\ \ m+n \geq 9\ \ \Rightarrow\ \ m \geq 5\ \ \text{or}\ \ n \geq 5\)

\(\text{Contrapositive statement:}\)

\(\text{If}\ \ m \ngeq 5\ \ \text{or}\ \ n \ngeq 5\ \ \Rightarrow\ \ \ m+n \ngeq 9\)

\(\text{i.e., if}\ \ m \lt 5\ \ \text{and}\ \ n \lt 5\ \ \Rightarrow\ \ \ m+n \lt 9\)

\(\Rightarrow C\)

Proof, EXT2 P1 2024 SPEC1 2

Prove that if \(x\) is an odd integer then \(2 x^2-3 x-7\) is even. (3 marks)

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\(\text{See worked solutions}\)

Show Worked Solution

\(\text{Let}\ \ x=2k+1,\ \ k \in Z\ \ (x\ \text{is odd)}\)

\(2 x^2-3 x-7\)	\(=2(2k+1)^2-3(k+1)-7\)
	\(=2(4k^2+4k+1)-3(2k+1)-7\)
	\(=8k^2+8k+2-6k-10\)
	\(=8k^2+2k-8\)
	\(=2(4k^2+k-4)\ \ \text{(which is even)}\)

Proof, EXT2 P1 2024 HSC 14d

The following argument attempts to prove that \(0=1\).

We evaluate \(\displaystyle\int \frac{1}{x}\, d x\) using the method of integration by parts.

	\(\displaystyle \int \frac{1}{x}\,d x\)	\(=\displaystyle \int \frac{1}{x} \times 1\, d x\)
		\(=\displaystyle\frac{1}{x} \times x-\int-\frac{1}{x^2} x\, d x\)
		\(=1+\displaystyle\int \frac{1}{x}\, d x\)

We may now subtract \(\displaystyle \int \frac{1}{x}\,d x\) from both sides to show that \(0=1\).

Explain what is wrong with this argument. (2 marks)

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\(\text {Consider the line in proof }\)

\(\displaystyle {\int \frac{1}{x}\, d x=1+\int \frac{1}{x}\, d x}\)

\(\Rightarrow \text{\(\ \)Each integral will have its own constant}\)

\(\Rightarrow \text{ If constants are \(c_1\) and \(c_2, \abs{c_1-c_2}=1\) in this proof.}\)

\(\Rightarrow \text{ Subtracting}\ \displaystyle \int \frac{1}{x}\,d x\ \text{from both sides invalidates the proof.}\)

Show Worked Solution

\(\text {Consider the line in proof }\)

\(\displaystyle {\int \frac{1}{x}\, d x=1+\int \frac{1}{x}\, d x}\)

♦♦ Mean mark 34%.

\(\Rightarrow \text{\(\ \)Each integral will have its own constant}\)

\(\Rightarrow \text{ If constants are \(c_1\) and \(c_2, \abs{c_1-c_2}=1\) in this proof.}\)

\(\Rightarrow \text{ Subtracting}\ \displaystyle \int \frac{1}{x}\,d x\ \text{from both sides invalidates the proof.}\)

Proof, EXT2 P1 2024 HSC 14a

Prove that if \(a\) is any odd integer, then \(a^2-1\) is divisible by 8. (2 marks)

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\(\text {Prove if \(a\) is odd, \(a^2-1\) is divisible by 8.}\)

\(\text {Let}\ \ a=2 n+1, \ \ n \in \mathbb{Z}\)

	\(a^2-1\)	\(=(2 n+1)^2-1\)
		\(=4 n^2+4 n+1-1\)
		\(=4 n(n+1)\)

\(\text{If \(n\) is odd (\(n=2 k+1, k \in Z)\):}\)

\(a^2-1=4(2 k+1)(2 k+2)=8(2 k+1)(k+1) /8\)

\(\text{If \(n\) is even \((n=2 k)\):}\)

\(a^2-1=4(2 k)(2 k+1)=8 k(2 k+1) /8\)

\(\therefore \text{ If \(a\) is odd, \(a^2-1\) is divisible by 8}\)

Show Worked Solution

\(\text {Prove if \(a\) is odd, \(a^2-1\) is divisible by 8.}\)

\(\text {Let}\ \ a=2 n+1, \ \ n \in \mathbb{Z}\)

	\(a^2-1\)	\(=(2 n+1)^2-1\)
		\(=4 n^2+4 n+1-1\)
		\(=4 n(n+1)\)

\(\text{If \(n\) is odd (\(n=2 k+1, k \in Z)\):}\)

\(a^2-1=4(2 k+1)(2 k+2)=8(2 k+1)(k+1) /8\)

\(\text{If \(n\) is even \((n=2 k)\):}\)

\(a^2-1=4(2 k)(2 k+1)=8 k(2 k+1) /8\)

\(\therefore \text{ If \(a\) is odd, \(a^2-1\) is divisible by 8}\)

Proof, EXT2 P1 2024 HSC 12d

Explain why there is no integer \(n\) such that \((n+1)^{41}-79 n^{40}=2\). (2 marks)

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\(\text{Show}\ \ \nexists \, n \in \mathbb{Z}: \ (n+1)^{41}-79 n^{40}=2\)

\(\text {If \(n\) is even:}\)

\(\text{LHS}=(\text{odd})^{41}-79(\text{even})^{40}=\text {odd}-\text {even}=\text {odd} \neq 2\)

\(\text {If \(n\) is odd:}\)

\(\text {LHS }=(\text {even})^{41}-79(\text {odd})^{40}=\text {even}- \text {odd}=\text {odd} \neq 2\)

\(\therefore\ \text {By contradiction}\)

\(\nexists \, n \in \mathbb{Z}: \ (n+1)^{41}-79 n^{40}=2\)

Show Worked Solution

\(\text{Show}\ \ \nexists \, n \in \mathbb{Z}: \ (n+1)^{41}-79 n^{40}=2\)

\(\text {If \(n\) is even:}\)

\(\text{LHS}=(\text{odd})^{41}-79(\text{even})^{40}=\text {odd}-\text {even}=\text {odd} \neq 2\)

\(\text {If \(n\) is odd:}\)

\(\text {LHS }=(\text {even})^{41}-79(\text {odd})^{40}=\text {even}- \text {odd}=\text {odd} \neq 2\)

\(\therefore\ \text {By contradiction}\)

\(\nexists \, n \in \mathbb{Z}: \ (n+1)^{41}-79 n^{40}=2\)

♦ Mean mark 49%.

Proof, EXT2 P1 2024 HSC 3 MC

Consider the statement:

'If a polygon is a square, then it is a rectangle.'

Which of the following is the converse of the statement above?

If a polygon is a rectangle, then it is a square.
If a polygon is a rectangle, then it is not a square.
If a polygon is not a rectangle, then it is not a square.
If a polygon is not a square, then it is not a rectangle.

Show Answers Only

\(A\)

Show Worked Solution

\(\text{Statement:}\ P \Rightarrow \ Q\)

\(\text{Converse of statement:}\ Q \Rightarrow \ P\)

\(\Rightarrow A\)

Proof, EXT2 P1 2024 HSC 2 MC

Consider the following statement written in the formal language of proof

\(\forall \theta \in\biggl(\dfrac{\pi}{2}, \pi\biggr) \exists\ \phi \in\biggl(\pi, \dfrac{3 \pi}{2}\biggr) ; \ \sin \theta=-\cos \phi\).

Which of the following best represents this statement?

There exists a \(\theta\) in the second quadrant such that for all \(\phi\) in the third quadrant \(\sin \theta=-\cos \phi\).
There exists a \(\phi\) in the third quadrant such that for all \(\theta\) in the second quadrant \(\sin \theta=-\cos \phi\).
For all \(\theta\) in the second quadrant there exists a \(\phi\) in the third quadrant such that \(\sin \theta=-\cos \phi\).
For all \(\phi\) in the third quadrant there exists a \(\theta\) in the second quadrant such that \(\sin \theta=-\cos \phi\).

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\(C\)

Show Worked Solution

\(\Rightarrow C\)

Proof, EXT2 P1 2023 SPEC2 1 MC

Consider the following statement.

'If my football team plays badly, then they are not training enough.'

Which one of the following statements is the contrapositive of the statement above?

If my football team plays badly, then they need more training.
If they are training enough, then my football team does not play badly.
If my football team doesn't play badly, then they are training enough.
If they are training enough, then my football team will most likely win.

Show Answers Only

\( B\)

Show Worked Solution

\(\text{Statement is conditional:}\ X \Rightarrow Y \)

\(\text{Logically equivalent contrapositive statement:}\ \neg Y \Rightarrow \neg X\)

\(\Rightarrow B\)

Proof, EXT2 P1 2023 HSC 4 MC

Consider the following statement about real numbers.

"Whichever positive number \(r\) you pick, it is possible to find a number \(x\) greater than 1 such that

\(\dfrac{\ln x}{x^3}<r\). "

When this statement is written in the formal language of proof, which of the following is obtained?

\(\forall x>1 \quad \exists r>0 \quad \dfrac{\ln x}{x^3}<r\)
\(\exists x>1 \quad \forall r>0 \quad \dfrac{\ln x}{x^3}<r\)
\(\forall r>0 \quad \exists x>1 \quad \dfrac{\ln x}{x^3}<r\)
\(\exists r>0 \quad \forall x>1 \quad \dfrac{\ln x}{x^3}<r\)

Show Answers Only

\(C\)

Show Worked Solution

\(\text{Whichever positive number}\ r\ \text{you pick …}\ \forall r>0 \)

\(\text{It is possible to find a number}\ x\ \text{greater then 1 …}\ \exists x>1 \)

\(\text{Such that}\ \ \dfrac{\ln x}{x^3}<r\)

\(\Rightarrow C\)

Proof, EXT2 P1 2023 HSC 12a

Prove that \(\sqrt{23}\) is irrational. (3 marks)

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\(\text{Proof (See Worked Solutions)} \)

Show Worked Solution

\(\text{Proof by contradiction:} \)

\(\text{Assume}\ \sqrt{23}\ \text{is rational} \)

\( \sqrt{23} = \dfrac{p}{q}\ \ \text{where}\ p, q \in \mathbb{Z}\ \ \text{with no common factor except 1} \)

\(23\)	\(= \dfrac{p^2}{q^2} \)
\(23q^2\)	\(=p^2\)

\(\Rightarrow \text{23 is a factor of}\ p^2 \)

\(\Rightarrow \text{23 is a factor of}\ p \)

\( \exists k \in \mathbb{Z}\ \ \text{such that}\ \ p=23k \)

\(23q^2\)	\(=(23k)^2 \)
\(q^2\)	\(=23k^2 \)

\(\Rightarrow \text{23 is a factor of}\ q^2 \)

\(\Rightarrow \text{23 is a factor of}\ q \)

\(\therefore \text{HCF}\ \geq 23 \)

\(\therefore \text{By contradiction,}\ \sqrt{23}\ \text{is rational} \)

Proof, EXT2 P1 2023 HSC 2 MC

Consider the following statement.

'If an animal is a herbivore, then it does not eat meat.'

Which of the following is the converse of this statement?

If an animal is a herbivore, then it eats meat.
If an animal is not a herbivore, then it eats meat.
If an animal eats meat, then it is not a herbivore.
If an animal does not eat meat, then it is a herbivore.

Show Answers Only

\(D\)

Show Worked Solution

\(\text{Statement:}\ \ P \Rightarrow Q \)

\(\text{Converse of statement:}\ \ Q \Rightarrow P \)

\(\Rightarrow D\)

Proof, EXT2 P1 2022 HSC 13a

Prove that for all integers `n` with `n >= 3`, if `2^(n)-1` is prime, then `n` cannot be even. (3 marks)

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text{Contrapositive statement:}`

`text{If}\ n\ text{is even}, 2^n-1\ text{is NOT prime.}`

`text{Let}\ \ n=2k,\ \ (kinZZ and k>=2)`

`2^n-1`	`=2^(2k)-1`
	`=(2^k)^2-1`
	`=(2^k-1)(2^k+1)`

`text{S}text{ince}\ \ k>=2\ \ =>\ \ 2^k-1>=3 and 2^k+1>=5`

`:.2^n-1\ \ text{is not prime if}\ n\ text{is even, as it has two non-trivial integer factors.}`

`:.\ text{By contrapositive statement, if}\ 2^(n)-1\ text{is prime}, n\ text{cannot be even.}`

Proof, EXT2 P1 2022 HSC 7 MC

Consider the statement `P`.

`P` : For all integers `n \geq 1`, if `n` is a prime number then `(n(n+1))/(2)` is a prime number.

Which of the following is true about this statement and its converse?

The statement `P` and its converse are both true.
The statement `P` and its converse are both false.
The statement `P` is true and its converse is false.
The statement `P` is false and its converse is true.

Show Answers Only

`D`

Show Worked Solution

`text{Statement:}\ \ ∀n in ZZ^+, text{if}\ n\ text{is prime}\ \ =>\ \ (n(n+1))/(2)\ text{is prime.}`

`text{Converse:}\ \ ∀n in ZZ^+, text{if}\ (n(n+1))/(2)\ text{is prime}\ \ =>\ \ n\ text{is prime.}`

`text{Consider prime}\ n=3:`

`(n(n+1))/(2)=(3xx4)/2=6\ \ text{(not prime → Statement is false)}`

`text{Consider}\ \ (n(n+1))/(2):`

`(n(n+1))/(2)\ text{is a prime}\ iff\ n=2\ \ text{(prime)}`

`:.\ text{Converse is true}`

`=>D`

♦♦ Mean mark 37%.

Proof, EXT2 P1 2022 HSC 3 MC

Let `A, B, P` be three points in three-dimensional space with `A \ne B`.

Consider the following statement.

If `P` is on the line `A B`, then there exists a real number `\lambda` such that `vec{A P}=\lambda \vec{A B}`.

Which of the following is the contrapositive of this statement?

If for all real numbers `\lambda, \vec{A P}=\lambda \vec{A B}`, then `P` is on the line `A B`.
If for all real numbers `\lambda, \vec{A P} \ne \lambda \vec{A B}`, then `P` is not on the line `A B`.
If there exists a real number `\lambda` such that `\vec{A P}=\lambda \vec{A B}`, then `P` is on the line `A B`.
If there exists a real number `\lambda` such that `\vec{A P} \ne \lambda \vec{A B}`, then `P` is not on the line `A B`.

Show Answers Only

`B`

Show Worked Solution

`text{Statement:}`

`text{If}\ P\ text{is on}\ AB\ \ =>\ \ EE lambda\ \ text{such that}\ \ vec{A P}=\lambda \vec{A B}`

`text{Contrapositive statement:}`

`text{If}\ not \ EE lambda\ \ text{such that}\ \ vec{A P}=\lambda \vec{A B}\ \ =>\ \ P\ text{is not on}\ AB`

`text{In other words …}`

`text{If for all real numbers}\ lambda, \vec{A P} \ne \lambda \vec{A B}, text{then}\ P\ text{is not on the line}\ A B.`

`=>B`

Proof, EXT2 P1 2022 HSC 2 MC

The following proof aims to establish that `-4 = 0`

`text{Let}`	`a=-4`
`=>`	`a^2 = 16 \ text{and} `	`\ 4a + 4 = -12`	`text{Line 1}`
`=>`	`a^2 + 4a + 4 =`	`4`	`text{Line 2}`
`=>`	`(a + 2)^2 =`	`2^2`	`text{Line 3}`
`=>`	`a + 2 =`	`2`	`text{Line 4}`
`=>`	`a =`	`0`

At which line is the implication incorrect?

Line 1
Line 2
Line 3
Line 4

Show Answers Only

`D`

Show Worked Solution

`text{Consider Line 4:}`

`text{Given}\ \ (a + 2)^2=2^2\ \ =>\ \ a+2=+-2`

`=>D`

Proof, EXT2 P1 2021 HSC 15d

Prove that `2^n + 3^n != 5^n` for all integers `n ≥ 2`. (2 marks)

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`text{See Worked Solution}`

Show Worked Solution

`5^n= (2 + 3)^n`

`text(Expanding)\ \ (2 + 3)^n:`

Mean mark 51%.

`5^n= 2^n +\ ^n C_1 *2^{n-1} *3 + \ … \ + \ ^n C_{n-1} *2 * 3^{n-1} + 3^n`

`text{S} text{ince} \ \ ^n C_1 * 2^{n-1} *3 + \ … \ + \ ^n C_{n-1} * 2 * 3^{n-1} > 0 \ \ text{for} \ n ≥ 2`

`=> 5^n > 2^n + 3^n \ \ text{for} \ n ≥ 2`

`:. 5^n ≠ 2^n + 3^n \ \ text{for} \ n ≥ 2`

Proof, EXT2 P1 2021 HSC 15b

For integers `n ≥ 1`, the triangular numbers `t_n` are defined by `t_n = (n(n + 1))/2`, giving `t_1 = 1, t_2 = 3, t_3 = 6, t_4 = 10` and so on.

For integers `n >= 1`, the hexagonal numbers `h_n` are defined by `h_n = 2n^2-n`, giving `h_1 = 1, h_2 = 6, h_3 = 15, h_4 = 28` and so on.

Show that the triangular numbers `t_1, t_3 , t_5`, and so on, are also hexagonal numbers. (2 marks)

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Show that the triangular numbers `t_2, t_4 , t_6`, and so on, are not hexagonal numbers. (1 mark)

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i. `text(See Worked Solution)`

ii. `text(See Worked Solution)`

Show Worked Solution

i. `t_n = (n(n + 1))/2`

♦ Mean mark (i) 50%.

`text(Odd triangular numbers:)`

`→ \ text(Let)\ \ n = 2k-1\ \ text(for integers)\ \ k >= 1`

`t_(2k-1)`	`= ((2k-1)(2k-1 + 1))/2`
	`= (2k(2k-1))/2`
	`= 2k^2-k\ \ text{(definition of hexagonal numbers)}`

`:. t_1, t_3\ …\ text(are also hexagonal numbers.)`

ii. `t_2, t_4, t_6, …`

♦ Mean mark (ii) 1%!

`→\ text(Let)\ \ n = 2k\ \ text(for integers)\ \ k > = 1`

`t_(2k)`	`= (2k(2k + 1))/2`
	`= 2k^2 + k`

`text(Find values for)\ \ k, n\ \ text(that satisfy:)`

`2k^2 + k`	`= 2n^2-n`
`2k^2-2n^2 + k + n`	`= 0`
`2(k-n)(k + n) + (k + n)`	`= 0`
`(k + n)(2k-2n + 1)`	`= 0`

`k = -n =>\ text(no solution)\ (k, n >= 1)`

`2k = 2n-1 =>\ text(no solution)\ \ (2k\ \ text(is even,)\ \ 2n-1\ \ text{is odd)}`

`:. t_2, t_4, t_6, …\ text(are not hexagonal numbers.)`

Proof, EXT2 P1 2021 HSC 12b

Consider Statement A.

Statement A: ‘If `n^2` is even, then `n` is even.’

What is the converse of Statement A?. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Show that the converse of Statement A is true. (1 mark)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`text{See Worked Solution}`
`text{See Worked Solution}`

Show Worked Solution

i. `text{Converse}`

`text{If} \ n \ text{is even, then} \ n^2 \ text{is even.}`

ii. `text{If} \ n \ text{is even:}`

`n`	`= 2p, \ p ∈ ZZ`
`n^2`	`= (2 p)^2`
	`=4 p^2`
	`= 2 (2p^2)`
	`=2q, \ q ∈ ZZ`

`:. \ text{If} \ n \ text{is even, then} \ n^2 \ text{is even.}`

Proof, EXT2 P1 2021 HSC 9 MC

Four cards have either RED or BLACK on one side and either WIN or TRY AGAIN on the other side.

Sam places the four cards on the table as shown below.

A statement is made: ‘If a card is RED, then it has WIN written on the other side’.

Sam wants to check if the statement is true by turning over the minimum number of cards.

Which cards should Sam turn over?

1 and 4
3 and 4
1. 2 and 4
1, 3 and 4

Show Answers Only

`B`

Show Worked Solution

`text{L} text{ogically equivalent statements are:}`

`text{Red}`	`=> \ text{Win} \ …\ (1)`
`¬ \ text{Win}`	`=> \ ¬ \ text{Red} \ …\ (2)`

`text{To confirm statement is true}`

♦ Mean mark 48%.

`text{Card 1 – no need to turn}`

`text{Card 2 – no need to turn}`

`text{Card 3 – turn to confirm (2)}`

`text{Card 4 – turn to confirm (1)}`

`=> B`

Proof, EXT2 P1 2021 HSC 5 MC

Which of the following statements is FALSE?

`∀ a, b ∈ RR,` `a \ a^3 < b^3`
`∀ a, b ∈ RR,` `a e^{-a} > e^{-b}`
`∀ a, b ∈ (0, + ∞),` `a \ text{ln} \ a < text{ln} \ b`
`∀ a, b ∈ RR, text{with} \ a,b ≠ 0,` `a \ 1/a > 1/b`

Show Answers Only

`D`

Show Worked Solution

`text{By contradiction:}`

♦ Mean mark 50%.

`text{Consider} \ D`

`text{Let} \ \ a = -1 \ \ text{and} \ \ b = 1,`

`a \ -1 < 1 \ \ text{(TRUE)}`

`1/a > 1/b \ -> \ -1 > 1 \ \ text{(FALSE)}`

`=>\ D \ text{is false}`

Proof, EXT2 P1 2021 HSC 4 MC

Consider the statement:

‘For all integers `n`, if `n` is a multiple of 6, then `n` is a multiple of 2’.

Which of the following is the contrapositive of the statement?

There exists an integer `n` such that `n` is a multiple of 6 and not a multiple of 2.
There exists an integer `n` such that `n` is a multiple of 2 and not a multiple of 6.
For all integers `n`, if `n` is not a multiple of 2, then `n` is not a multiple of 6.
For all integers `n`, if `n` is not a multiple of 6, then `n` is not a multiple of 2.

Show Answers Only

`C`

Show Worked Solution

`text{L} text{ogically equivalent statements:}`

`text{If} \ n \ text{is multiple of 6} \ => \ n \ text{is a multiple of 2}`

`¬\ n\ text{is a multiple of 2}\ => \ ¬\ n \ text{is a multiple of 6}`

`text{i.e. if} \ n \ text{is not multiple of 2} \ => \ n \ text{is not a multiple of 6}`

`=>\ C`

Proof, EXT2 P1 EQ-Bank 12

Prove `sqrt5 + sqrt3 > sqrt14` by contradiction. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

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`text(See Worked Solutions)`

Show Worked Solution

`text(Proof by contradiction:)`

`text(Assume)\ \ sqrt5 + sqrt3 <= sqrt14`

`(sqrt5 + sqrt3)^2`	`<= (sqrt14)^2`
`5 + 2sqrt15 + 3`	`<= 14`
`2sqrt15`	`<= 6`
`sqrt15`	`<= 3`
`15`	`<= 9\ \ \ text{(incorrect)}`

`:. text(By contradiction,)\ sqrt5 + sqrt3 > sqrt14`

Proof, EXT2 P1 EQ-Bank 23

Prove that `log_3 7` is irrational. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

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`text(See Worked Solution)`

Show Worked Solution

`text(Proof by contradiction:)`

`text(Assume that)\ \ log_3 7\ \ text(is rational.)`

`text(i.e.)\ \ log_3 7 = p/q\ text(where)\ \ p, q ∈ ZZ\ \ text(with no common factor except 1.)`

`log_3 7`	`= p/q`
`q log_3 7`	`= p`
`log_3 7^q`	`= p`
`7^q`	`= 3^p`

`text(S)text(ince 7 and 3 are prime numbers), 7^q != 3^p`

`:.\ text(Contradiction: integer values)\ \ p, q\ \ text(do not exist.)`

`:. log_3 7\ text(is irrational.)`

Proof, EXT2 P1 2020 HSC 8 MC

Consider the statement:

"If `n` is even, then if `n` is a multiple of 3, then `n` is a multiple of 6".

Which of the following is the negation of this statement?

`n` is odd and `n` is not a multiple of 3 or 6.
`n` is even and `n` is a multiple of 3 but not a multiple of 6.
If `n` is even, then `n` is not a multiple of 3 and `n` is not a multiple of 6.
If `n` is odd, then if `n` is not a multiple of 3 then `n` is not a multiple of 6.

Show Answers Only

`B`

Show Worked Solution

`text{Proposition: If} \ \ X =>Y`

♦♦ Mean mark 33%.

`X \ text{is a compound statement}`

`text{“If} \ n \ text{is even and a multiple of 3.”}`

`Y \ text{states “} n \ text{is a multiple of 6.”}`

`text{Negation if} \ X \ text{but} \ ¬ \ Y.`

`=> \ B`

Proof, EXT2 P1 2020 HSC 14d

Prove that for any integer `n > 1, log_n (n + 1)` is irrational. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

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`text{See Worked Solutions}`

Show Worked Solution

`text{Proof by contradiction:}`

`text{Assume} \ log_n(n + 1) \ text{is rational}`

`therefore \ log_n (n + 1) = frac{p}{q} \ \ text{where} \ \ p,q ∈ ZZ \ text{with no common factor except 1}`

`n^(frac{p}{q}`	`= n + 1`
`n^p`	`= (n + 1)^q`

`text{Strategy 1}`

`n^p = (n + 1)^q \ \ text{when} \ \ p = q = 0\ \ text{only}`

`q ≠ 0`

`:.\ text{By contradiction}, log_n (n + 1) \ \ text{is irrational.}`

`text{Strategy 2}`

`n^p = (n + 1)^q`

`text{If} \ \ n\ \ text{is odd, LHS is odd and RHS is even.}`

`text{If} \ \ n\ \ text{is even LHS is even and RHS is odd.}`

`text{Statement is true for} \ \ p = q = 0 , text{but} \ \ q ≠ 0`

`therefore \ text{By contradiction,} \ log_n (n + 1) \ text{is irrational.}`

Proof, EXT2 P1 2020 HSC 15a

In the set of integers, let `P` be the proposition:

'If `k + 1` is divisible by 3, then `k^3 + 1` divisible by 3.'

Prove that the proposition `P` is true. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Write down the contrapositive of the proposition `P`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Write down the converse of the proposition `P` and state, with reasons, whether this converse is true or false. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

`text{See Worked Solutions}`
`text{See Worked Solutions}`
`text{See Worked Solutions}`

Show Worked Solution

i. `text{Let} \ \ k + 1 = 3N, \ N∈ Z`

`=> k = 3N-1`

`k^3 + 1`	`= (3N-1)^3 + 1`
	`= (3N)^3 + 3(3N)^2 (-1) + 3(3N)(-1)^2 + (-1)^3 + 1`
	`= 27N^3-27N^2 + 9N-1 + 1`
	`= 3 (9N^3-9N^2 + 3N)`
	`= 3Q \ , \ Q ∈ Z`

`therefore \ text{If} \ \ k+ 1 \ \ text{is divisible by 3}, text{then} \ \ k^3 + 1 \ \ text{is divisible by 3.}`

ii. `text{Contrapositive}`

`text{If} \ \ k^3 + 1 \ \ text{is not divisible by 3, then}\ \ k + 1\ \ text{is not divisible by 3.}`

♦♦ Mean mark (iii) 36%.

iii. `text{Converse:}`

`text{If} \ \ k^3 + 1\ \ text{is divisible by 3, then}\ \ k + 1\ \ text{is divisible by 3.}`

`text(Contrapositive of converse:)`

`text{If}\ \ k + 1\ \ text{is not divisible by 3, then}\ \ k^3 + 1\ \ text{is not divisible by 3.}`

`text(i.e.)\ \ k + 1 \ \ text{is not divisible by 3 when}\ \ k + 1 = 3Q + 1\ \ text{or}\ \ k + 1 = 3Q + 2, text{where}\ Q ∈ Z`

`text{If} \ \ k + 1`	`= 3Q + 1\ \ => \ k=3Q`
`k^3 + 1`	`= (3Q)^3 + 1`
	`= 27Q^3 + 1`
	`= 3(9Q^3) + 1`
	`= 3M + 1 \ \ (text{not divisible by 3,}\ M ∈ Z)`

`text{If} \ \ k + 1`	`= 3Q + 2\ \ => \ k=3Q+1`
`k^3 + 1`	`= (3Q + 1)^3 + 1`
	`= (3Q)^3 + 3(3Q)^2 + 3(3Q) + 1 + 1`
	`= 27Q^3 + 27Q^2 + 9Q + 2`
	`= 3(9Q^3 + 9Q^2 + 3Q) + 2`
	`= 3M + 2 \ (text{not divisible by 3,}\ M ∈ Z) `

`therefore \ text{By contrapositive, if}\ \ k^3 + 1\ \ text {is divisible by 3, k + 1 is divisible by 3.}`

Proof, EXT2 P1 2020 HSC 7 MC

Consider the proposition:

'If `2^n-1` is not prime, then `n` is not prime'.

Given that each of the following statements is true, which statement disproves the proposition?

`2^5-1` is prime
`2^6-1` is divisible by 9
`2^7-1` is prime
`2^11-1` is divisible by 23

Show Answers Only

`D`

Show Worked Solution

`text(Strategy 1 – Contradiction)`

`text(Consider option)\ D,`

`text(S)text(ince)\ \ 2^11 -1\ \ text(is divisible by 23, it is NOT prime.)`

`text(The proposition states that 11 is not prime which is false.)`

`:. 2^11-1\ \ text(is divisible by 23, disproves the proposition.)`

`text(Strategy 2 – Contrapositive)`

`text{The proposition is conditional}`

`X => Y`

`text{L}text{ogically equivalent contrapositive statement}`

`not \ Y => not \ X`

`text{i.e. If} \ n \ text{is prime} \ => \ 2^n-1 \ text{is prime.}`

`text{Consider D:}`

`n = 11 \ text{(prime)}`

`2^11-1 \ text{is divisible by 23 (not prime)}`

`therefore \ text{Contrapositive statement is false and disproves the proposition.}`

`=> \ D`

Proof, EXT2 P1 EQ-Bank 14

If `(n-3)^2` is an even integer, prove by contrapositive that `n` is odd. (2 marks)

--- 8 WORK AREA LINES (style=lined) ---

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Statement)`

`text(If) \ \ (n-3)^2 \ \ text(is even) => n \ text(is odd)`

`text(Contrapositive)`

`text(If) \ \ n \ not \ text(odd) => (n-3)^2 \ not \ text(even)`

`text{(i.e.)}\ \ \ n \ text(even) => (n-3)^2 \ text(is odd)`

`text(If)\ n\ text(even), \ ∃ \ k, \ k ∈ Ζ \ \ text(where)\ \ n = 2k`

`(n-3)^2`	`= (2k-3)^2`
	`= 4k^2-12k + 9`
	`= 4(k^2-12k + 2) + 1`

`=> (n-3)^2 \ text(is odd)`

`:. \ text(If)\ n\ text(is even), (n-3)^2\ text(is odd)`

`:. \ text(By contrapositive, statement is true.)`

Proof, EXT2 P1 EQ-Bank 25

If `ab` is divisible by 3, prove by contrapositive that `a` or `b` is divisible by 3. (3 marks)

--- 12 WORK AREA LINES (style=lined) ---

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Statement)`

`ab \ text(is divisible by 3)\ => a \ text(or) \ b \ text(is divisible by) \ 3`

`text(Contrapositive)`

`a \ text(or) \ b not \ text(divisible by 3)\ => ab not \ text(divisible by 3)`

`text(Let) \ \ a`	`= 3x + p, \ text(where) \ \ x ∈ Ζ \ \ text(and) \ \ p= 1 \ text(or) \ \ 2`
`b`	`= 3y + q, \ text(where) \ \ y ∈ Ζ \ \ text(and) \ \ q= 1 \ text(or) \ \ 2`

`ab`	`= (3x + p)(3y + q)`
	`= 9xy + 3qx + 3py + pq`
	`= 3(3xy + qx + py) + pq`

`text(Possible values of) \ \ pq = 1, 2, 4`

`=> \ pq \ text(is not divisible by 3)`

`:. \ ab \ text(is not divisible by 3)`

`:. \ text(By contrapositive, statement is true.)`

Proof, EXT2 P1 EQ-Bank 22

If `a^2-4a + 3` is even, `a ∈ Ζ`,

prove by contrapositive that `a` is odd. (3 marks)

--- 9 WORK AREA LINES (style=lined) ---

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Proof by contrapositive)`

`a not \ text(odd)`	`\ =>\ \ a^2-4x + 3 not \ text(even)`
`text{(i.e) If} \ a\ text(is even) `	`\ => \ a^2-4x + 3 \ \ text(is odd)`

`text(If) \ a \ text(is even) , ∃ \ k , k ∈ Ζ , text(such that) \ \ a = 2k`

`text(Substitute) \ \ 2k \ \ text(into) \ \ a^2-4x + 3`

`(2k)^2-4(2k) + 3`	`= 4k^2-8k + 3`
	`= 2(2k^2-4k + 1) + 1`

`:. \ text(By contrapositive, if) \ \ a^2-4x + 3 \ \ text(is even) \ => \ a \ text(is odd.)`

Proof, EXT2 P1 EQ-Bank 11

Prove that `sqrt11 - sqrt5 < sqrt2` by contradiction. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Proof by contradiction:)`

`text(Assume that)\ \ sqrt11 – sqrt5 >= sqrt2`

`( sqrt11 – sqrt5)^2`	`>= 2`
`11 – 2 sqrt55 + 5`	`>= 2`
`2 sqrt55`	`<= 14`
`sqrt55`	`<= 7`
`55`	`<= 49 \ \ text{(which is incorrect)}`

`:.\ text(By contradiction,) \ sqrt11 – sqrt5 < sqrt2`

Proof, EXT2 P1 EQ-Bank 5 MC

Four cards are placed on a table with a letter on one face and a shape on the other.

You are given the rule: "if N is on a card then a circle is on the other side."

Which cards need to be turned over to check if this rule holds?

N and G
G and triangle
circle and N
N and triangle

Show Answers Only

`=> \ D`

Show Worked Solution

`text(Solution 1)`

`text(L)text(ogically equivalent statements are:)`

`N`	`=> \ text(circle)`
`not\ text(circle)`	`=> not N`

`text(To confirm rule is not broken,)`

`N \ text(must be turned)`

`text{Triangle (not circle) must be turned – only other shape not a circle.}`

`=> D`

`text(Solution 2)`

`text(Consider the flip side of each card.)`

`text(If circle has) \ N \ text(on the other side (or not)) – text(tells us nothing.)`

`text(If) \ G \ text(has a circle on the other side (or not)) – text(tells us nothing.)`

`text(If) \ N \ text(doesn’t have a circle on other side) – text(rule broken.)`

`text(If triangle has an) \ N \ text(on other side) – text(rule broken.)`

`:. \ text(Need to turn) \ N \ text(and triangle)`

Proof, EXT2 P1 EQ-Bank 2 MC

Consider the following statement.

"If you have no treasure, I have no kingdom."

Which of the following is logically equivalent to this statement?

If I have no kingdom then you have no treasure.
If you have treasure then I have a kingdom.
If you have no kingdom then I have no treasure.
If I have a kingdom then you have treasure.

Show Answers Only

`D`

Show Worked Solution

`text(The statement is conditional.)`

`text(If)\ X\ text(then)\ Y\ \ text(or)\ \ X=>Y`

`text(The contrapositive statement is logically equivalent.)`

`text{(i.e.)}\ \ ¬Y => ¬X`

`text(If I have a kingdom then you have treasure.)`

`=> D`

Proof, EXT2 P1 EQ-Bank 26

Prove that `log_2 11` is irrational. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Proof by contradiction:)`

`text(Assume that)\ \ log_2 11\ \ text(is rational.)`

`:. log_2 11 = p/q\ \ \ text(where)\ \ p,q in ZZ\ \ text(with no common factor except 1)`

`qlog_2 11`	`=p`
`log_2 11^q`	`=p`
`11^q`	`=2^p`

`text(LHS is odd.)`

`text(RHS is even.)`

`:.\ text(Contradiction: integer values)\ \ p, q\ \ text(do not exist.)`

`:.\ log_2 11\ \ text(is irrational.)`

Proof, EXT2 P1 EQ-Bank 29

If `n` is a positive integer,

prove `sqrt(10n+2)` is always irrational. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Proof by contradiction:)`

`text(Assume that)\ \ sqrt(10n+2)\ \ text(is rational.)`

`:. sqrt(10n+2) = p/q\ \ \ text(where)\ \ p,q in ZZ\ \ text(with no common factor except 1)`

`10n+2`	`=p^2/q^2`
`2q^2(5n+1)`	`=p^2\ …\ (1)`

`p^2\ \ text(is even) \ =>\ p\ text(is even)`

`text{(i.e.)}\ \ ∃ k,\ \ k in ZZ\ \ text(such that)\ \ p=2k`

`text{Substitute}\ \ p=2k\ \ text{into (1)}`

`2q^2(5n+1)`	`=(2k)^2`
`q^2(5n+1)`	`=2k^2`

`q^2(5n+1)\ \ text(is even since) \ =>\ 2k^2\ text(is even)`

`=> q^2\ \ text(is even since)\ \ 5n + 1\ \ text(can be odd or even).`

`=> q\ \ text(is even)`

`=> \ text(Contradiction:)\ p and q\ \ text(have a common factor of 2)`

`:.\ sqrt(10n+2)\ \ \ text(is irrational.)`

Proof, EXT2 P1 EQ-Bank 27

Prove that `root3 2` is irrational. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Proof by contradiction:)`

`text(Assume that)\ \ root3 2\ \ text(is rational.)`

`:. root3 2 = p/q\ \ \ text(where)\ \ p,q in ZZ\ \ text(with no common factor except 1)`

`2`	`=p^3/q^3`
`2q^3`	`=p^3\ …\ (1)`

`p^3\ \ text(is even) \ =>\ p\ text(is even)`

`text{(i.e.)}\ \ ∃ k,\ \ k in ZZ\ \ text(such that)\ \ p=2k`

`text{Substitute}\ \ q=2k\ \ text{into (1)}`

`2q^3`	`=(2k)^3`
`2q^3`	`=8k^3`
`q^3`	`=4k^3`

`q^3\ \ text(is even) \ =>\ q\ text(is even)`

`:. p and q\ \ text(have a common factor of 2)`

`=> \ text(Contradiction)`

`:.\ root3 2\ \ \ text(is irrational.)`

Proof, EXT2 P1 EQ-Bank 13

Prove that `1/sqrt2` is irrational. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Proof by contradiction:)`

`text(Assume that)\ \ 1/sqrt2\ \ text(is rational.)`

`:. 1/sqrt2 = p/q\ \ \ text(where)\ \ p,q in ZZ\ \ text(with no common factor except 1)`

`1/2`	`=p^2/q^2`
`q^2`	`=2p^2\ …\ (1)`

`q^2\ \ text(is even) \ =>\ q\ text(is even)`

`text{(i.e.)}\ \ ∃ k,\ \ k in ZZ\ \ text(such that)\ \ q=2k`

`text{Substitute}\ \ q=2k\ \ text{into (1)}`

`(2k)^2`	`=2p^2`
`2k^2`	`=p^2`

`p^2\ \ text(is even) \ =>\ p\ text(is even)`

`:. p and q\ \ text(have a common factor of 2)`

`=> \ text(Contradiction)`

`:.\ 1/sqrt2\ \ \ text(is irrational.)`

Proof, EXT2 P1 EQ-Bank 24

Prove that `sqrt3` is irrational. (3 marks)

--- 10 WORK AREA LINES (style=lined) ---

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Proof by contradiction:)`

`text(Assume that)\ \ sqrt3\ \ text(is rational.)`

`:. sqrt3 = p/q\ \ \ text(where)\ \ p,q in ZZ\ \ text(with no common factor except 1)`

`3`	`=p^2/q^2`
`3q^2`	`=p^2\ …\ (1)`

`text(If)\ \ q^2\ \ text(is even):`

COMMENT: `(2//q)` can be used for “2 divides `q`” or `q` is divisible by 2.

`=> q\ \ text(is even)\ \ (2//q)`

`=>p^2\ \ text(is even)\ \ =>\ p\ \ text(is even)\ \ (2//p)`

`p,q\ \ text(have a common factor of 2`

`text(Contradiction)\ =>\ sqrt3\ \ text(is irrational for)\ \ p, q\ \ text(even.)`

`text(If)\ \ q^2\ \ text(is odd):`

`=> 3q^2\ \ text(is odd)\ =>\ q\ \ text(is odd)`

`=>p^2\ \ text(is odd)\ \ =>\ p\ \ text(is odd)`

`text(Let)\ \ p=2x+1, \ \ q=2y+1,\ \ x,y inZZ`

`text{Substitute into (1)}`

`3(2y+1)^2`	`=(2x+1)^2`
`3(4y^2+4y+1)`	`=4x^2+4x+1`
`12y^2+12y+3`	`=4x^2+4x`
`6y^2+6y+1`	`=2x^2+2x`

`text(LHS) = 6(y^2+y)+1\ \ text(which is odd)`

`text(RHS) = 2(x^2+x)\ \ text(which is even)`

`text(Contradiction)\ =>\ sqrt3\ \ text(is irrational for)\ \ p, q\ \ text(odd.)`

`:.\ sqrt3\ \ \ text(is irrational.)`