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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:}\)

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

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

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

Proof, EXT2 P1 2023 HSC 12a

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Proof, EXT2 P1 2021 HSC 5 MC

Which of the following statements is FALSE?

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

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`text{By contradiction:}`

♦ Mean mark 50%.

`text{Consider} \ D`

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

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

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

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

Proof, EXT2 P1 SM-Bank 15

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

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`text(See Worked Solutions)`

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`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 SM-Bank 14

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

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`text(See Worked Solution)`

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`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 14d

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

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`text{See Worked Solutions}`

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`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 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?

  1. `2^5 - 1` is prime
  2. `2^6 - 1` is divisible by 9
  3. `2^7 - 1` is prime
  4. `2^11 - 1` is divisible by 23
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`D`

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`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 SM-Bank 10

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

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`text{Proof (See Worked Solutions)}`

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`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 SM-Bank 8

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

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`text{Proof (See Worked Solutions)}`

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`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 SM-Bank 9

If  `n`  is a positive integer,

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

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`text{Proof (See Worked Solutions)}`

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`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 SM-Bank 4

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

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`text{Proof (See Worked Solutions)}`

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`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 SM-Bank 3

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

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`text{Proof (See Worked Solutions)}`

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`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 SM-Bank 2

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

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`text{Proof (See Worked Solutions)}`

 

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`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.)`