Proof (deductive, induction, contradiction) – IB Maths exam questions with mark scheme

1StandardNon-calculatorHL only5 marks

Use proof by contradiction to show that $\sqrt3$ is irrational.

Assumes $\sqrt3=\dfrac pq$ where $p,q\in\mathbb{Z}$, $q\neq0$, with no common factor (allow $\gcd(p,q)=1$ / fraction in lowest terms)R1
Squares: $p^2=3q^2$M1
So $p^2$ is a multiple of 3, hence $p$ is a multiple of 3 (since 3 is prime; accept 'if $p$ is not a multiple of 3 then neither is $p^2$')R1
Sets $p=3m$: $9m^2=3q^2$, so $q^2=3m^2$ and $q$ is a multiple of 3M1
Both multiples of 3 contradicts no common factor, so $\sqrt3$ is irrationalR1
Answer: Proof (contradiction)
Examiner tip: Say why $p^2$ divisible by 3 implies $p$ divisible by 3, and state the contradiction explicitly.

2StandardNon-calculatorHL only6 marks

Prove by mathematical induction that, for all $n\in\mathbb{Z}^+$,

$$\sum_{r=1}^{n}r(r+1)=\frac{n(n+1)(n+2)}{3}.$$

$n=1$: LHS $=1\times2=2$, RHS $=\dfrac{1\times2\times3}{3}=2$, so true for $n=1$A1
Assumes true for $n=k$: $\sum_{r=1}^k r(r+1)=\dfrac{k(k+1)(k+2)}{3}$ (must be stated as an assumption; 'let $n=k$' alone does not earn this mark)M1
Considers $n=k+1$ and adds the next term $(k+1)(k+2)$M1
Correct intermediate step, e.g. $\dfrac{k(k+1)(k+2)+3(k+1)(k+2)}{3}$ or $(k+1)(k+2)\left(\dfrac k3+1\right)$ (accept expanded $\dfrac{k^3+6k^2+11k+6}{3}$)A1
$=\dfrac{(k+1)(k+2)(k+3)}{3}=\dfrac{(k+1)\big((k+1)+1\big)\big((k+1)+2\big)}{3}$, i.e. the formula with $n=k+1$A1
True for $n=1$, and true for $n=k\Rightarrow$ true for $n=k+1$, so true for all $n\in\mathbb{Z}^+$ by induction (award only if both M1 marks have been earned)R1
Answer: Proof by induction
Examiner tip: The final conclusion needs all three ideas: true for $n=1$, true for $k$ implies $k+1$, hence all $n$.

3CoreNon-calculator3 marks

Prove that, for every integer $n$, the expression $(2n+3)^2-(2n-1)^2$ is a multiple of $8$.

Attempts to expand both brackets: $4n^2+12n+9-(4n^2-4n+1)$ (allow one sign slip in expansion for M1; accept use of difference of two squares $[(2n+3)-(2n-1)][(2n+3)+(2n-1)]=4(4n+2)$)M1
Simplifies correctly to $16n+8$A1
Writes as $8(2n+1)$ and concludes: since $2n+1$ is an integer, the expression is a multiple of $8$ (AG). Do not award if the factor $8$ is not shown explicitly or the integer reason is missingR1
Answer: $(2n+3)^2-(2n-1)^2=16n+8=8(2n+1)$, a multiple of 8
Examiner tip: Showing it works for a few values of n is not a proof: you must factor out 8 and say the other factor is an integer.

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4StandardNon-calculator5 marks

(a) Show that, for $n\in\mathbb{Z}^+$, $$\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}.$$ [2]

(b) Hence find the exact value of $$\sum_{n=1}^{99}\frac{1}{\sqrt{n+1}+\sqrt{n}}.$$ [3]

(a) Multiplies numerator and denominator by $\sqrt{n+1}-\sqrt{n}$ (or starts from RHS and multiplies by $\dfrac{\sqrt{n+1}+\sqrt n}{\sqrt{n+1}+\sqrt n}$)M1
(a) Denominator simplified: $(\sqrt{n+1})^2-(\sqrt n)^2=(n+1)-n=1$, leading to $\sqrt{n+1}-\sqrt n$ (AG). Must see the denominator equal to $1$ explicitlyA1
(b) Uses (a) to write the sum as $(\sqrt2-\sqrt1)+(\sqrt3-\sqrt2)+\dots+(\sqrt{100}-\sqrt{99})$ — at least first two and last term seenM1
(b) Recognises cancellation (telescoping), leaving $\sqrt{100}-\sqrt1$M1
(b) $9$ caoA1
Answer: (a) proof; (b) $9$
Examiner tip: In 'show that' questions the denominator simplifying to 1 must be written out; jumping straight to the given answer loses the A1.

5StandardNon-calculator4 marks

Prove that the sum of the squares of any two consecutive odd numbers is always $2$ more than a multiple of $8$.

Represents two consecutive odd numbers algebraically, e.g. $2n+1$ and $2n+3$ for $n\in\mathbb{Z}$ (accept $2n-1$, $2n+1$). Do not accept $n$ and $n+2$ without stating $n$ oddM1
Squares and adds: $(2n+1)^2+(2n+3)^2$ expandedM1
$8n^2+16n+10$ (or $8n^2+2$ for $2n\pm1$)A1
Rewrites as $8(n^2+2n+1)+2$ (or $8n^2+2$) and states $n^2+2n+1$ (or $n^2$) is an integer, so the sum is 2 more than a multiple of 8R1
Answer: $(2n+1)^2+(2n+3)^2=8(n+1)^2+2$
Examiner tip: Using x and x+2 without saying x is odd proves nothing about odd numbers; define them as 2n+1 and 2n+3.

6CoreNon-calculatorHL only4 marks

Each statement below is false. For each one, give a counterexample, showing clearly why it disproves the statement.

(a) For every positive integer $n$, the number $n^2+n+41$ is prime. [2]

(b) For all real numbers $a$ and $b$, if $a^2>b^2$ then $a>b$. [2]

(a) Chooses a valid value, e.g. $n=40$ (giving $1681$) or $n=41$ (giving $1763$); accept any $n$ that worksA1
(a) Shows the value is not prime, e.g. $1681=41^2$ or $1763=41\times43$. A value without a factorisation scores A1 R0R1
(b) Chooses a valid pair, e.g. $a=-3,\ b=1$ (any $a,b$ with $a^2>b^2$ and $a\le b$)A1
(b) Verifies both parts: $a^2=9>1=b^2$ but $a=-3<1=b$R1
Answer: (a) e.g. $n=40$: $1681=41^2$; (b) e.g. $a=-3,\ b=1$
Examiner tip: A counterexample earns full marks only when you show it satisfies the hypothesis and fails the conclusion.

7StandardNon-calculatorHL only6 marks

Prove by mathematical induction that $25^n-3^n$ is divisible by $22$ for all $n\in\mathbb{Z}^+$.

Base case: $n=1$: $25-3=22$, which is divisible by 22A1
Assumes true for $n=k$: $25^k-3^k=22m$ for some $m\in\mathbb{Z}$ (must state 'assume'; do not accept 'let $n=k$')M1
Considers $n=k+1$: $25^{k+1}-3^{k+1}=25\cdot25^k-3\cdot3^k$M1
Rewrites to use the assumption, e.g. $25(25^k-3^k)+22\cdot3^k$ (or $25^k=22m+3^k$ substituted)A1
$=22(25m+3^k)$, a multiple of 22A1
Concluding statement: true for $n=1$, and true for $n=k$ implies true for $n=k+1$, so true for all $n\in\mathbb{Z}^+$. Only award if at least 4 of the previous marks are earnedR1
Answer: Proof by induction
Examiner tip: The R1 is only available if your conclusion explicitly links the base case and the inductive step.

8HardNon-calculatorHL only7 marks

(a) Show that the inequality $n!>3^n$ is false when $n=6$. [1]

(b) Prove by mathematical induction that $n!>3^n$ for all integers $n\ge7$. [6]

(a) $6!=720$ and $3^6=729$, and $720<729$A1
(b) Base case: $n=7$: $7!=5040>2187=3^7$A1
Assumes $k!>3^k$ for some integer $k\ge7$M1
Considers $(k+1)!=(k+1)\,k!$M1
$(k+1)\,k!>(k+1)\,3^k$ using the assumptionA1
Since $k\ge7$, $k+1>3$, so $(k+1)\,3^k>3\cdot3^k=3^{k+1}$, hence $(k+1)!>3^{k+1}$. The reason $k+1>3$ must be statedR1
Concluding statement: true for $n=7$, and true for $n=k$ ($k\ge7$) implies true for $n=k+1$, so true for all $n\ge7$. Award only if at least 4 of the (b) marks earnedR1
Answer: (a) $720<729$; (b) proof by induction
Examiner tip: In an inequality induction you must justify the comparison step (here k+1>3); a chain of '>' signs with no reason loses the R1.

9HardNon-calculatorHL only7 marks

Let $f(x)=x\,\mathrm{e}^{2x}$.

Prove by mathematical induction that, for all $n\in\mathbb{Z}^+$, $$f^{(n)}(x)=2^{\,n-1}(2x+n)\,\mathrm{e}^{2x},$$ where $f^{(n)}(x)$ denotes the $n$th derivative of $f(x)$.

Uses product rule for $n=1$: $f'(x)=\mathrm{e}^{2x}+2x\,\mathrm{e}^{2x}$M1
$=(2x+1)\mathrm{e}^{2x}$ and $2^{0}(2x+1)\mathrm{e}^{2x}$ from the formula, so true for $n=1$A1
Assumes true for $n=k$: $f^{(k)}(x)=2^{k-1}(2x+k)\mathrm{e}^{2x}$M1
Differentiates the assumed expression using the product rule: $f^{(k+1)}(x)=2^{k-1}\left[2\,\mathrm{e}^{2x}+2(2x+k)\mathrm{e}^{2x}\right]$M1
Simplifies: $2^{k-1}\cdot2\,(2x+k+1)\mathrm{e}^{2x}$A1
$=2^{k}\big(2x+(k+1)\big)\mathrm{e}^{2x}$, which is the formula with $n=k+1$A1
Concluding statement: true for $n=1$, and true for $n=k$ implies true for $n=k+1$, so true for all $n\in\mathbb{Z}^+$. Award only if at least 4 of the previous marks are earnedR1
Answer: Proof by induction
Examiner tip: Differentiate the assumed $k$th derivative, not $f(x)$ again, and write the final form so it visibly matches the formula with $n=k+1$.

10StretchNon-calculatorHL only8 marks

(a) Use proof by contradiction to show that $\log_2 5$ is irrational. [4]

(b) Use proof by contradiction to show that there are no positive integers $a$ and $b$ such that $a^2-b^2=10$. [4]

(a) Assumes $\log_2 5=\dfrac pq$ with $p,q\in\mathbb{Z}^+$ (positive since $\log_25>0$)M1
(a) Converts to $2^{p/q}=5$, hence $2^p=5^q$A1
(a) Parity (or prime factorisation) argument: $2^p$ is even (as $p\ge1$) but $5^q$ is odd; accept 'unique prime factorisation means the two sides cannot be equal'R1
(a) States the contradiction, so $\log_2 5$ is irrationalR1
(b) Assumes such $a,b$ exist and factorises: $(a-b)(a+b)=10$M1
(b) Notes $a-b$ and $a+b$ have the same parity, since their difference $2b$ (or sum $2a$) is evenR1
(b) Considers the factor pairs of 10 with $a+b>0$: $1\times10$ or $2\times5$ (accept negative pairs dismissed as $a+b>0$ forces $a-b>0$), each consisting of one odd and one even factorR1
(b) Contradiction stated: no factor pair has equal parity, so no such positive integers exist. Accept alternative: if both factors same parity then product is odd or a multiple of 4, but 10 is neitherR1
Answer: (a) $2^p=5^q$ impossible (even vs odd); (b) $(a-b)(a+b)=10$ needs same-parity factors, none exist
Examiner tip: In (a) many write $p/q$ but forget to justify $p\ge1$ so that $2^p$ is even; in (b) state why the two factors must share parity.

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