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A-Level Pure 4 · Edexcel IAL WMA14 · Topic pack

Topic pack · Proof by contradiction แพ็กฝึกเฉพาะหัวข้อ · การพิสูจน์โดยข้อขัดแย้ง

10 questions 10 ข้อ 55 marks 55 คะแนน 70 min 70 นาที Proof by contradiction, easy → hard

Ten questions on one topic, ordered easy → hard. The last three are full exam difficulty. Open the mark schemes only after a real attempt. สิบข้อหัวข้อเดียว เรียงง่าย → ยาก สามข้อสุดท้ายคือระดับข้อสอบจริง เปิดมาร์คสกีมหลังลองทำจริงเท่านั้น

⏱ 70:00 Scoreคะแนนรวม: –
Q1. 3 marks

Prove by contradiction that there is no largest even integer.

Mark schemeshow ▾
B1 assume, for contradiction, that there is a largest even integer $N$  M1 consider $N+2$: since $N=2k$, $N+2=2(k+1)$ is also even  A1 but $N+2>N$, contradicting that $N$ is the largest even integer, so no largest even integer exists ∎.
Q2. 4 marks

Prove by contradiction that if $n$ is an integer and $n^2$ is even, then $n$ is even.

Mark schemeshow ▾
B1 assume, for contradiction, that $n^2$ is even but $n$ is odd  M1 write $n=2k+1$ and expand: $n^2=4k^2+4k+1$  A1 $n^2=2(2k^2+2k)+1$, which is odd  A1 this contradicts $n^2$ being even, so $n$ must be even ∎.
Q3. 4 marks

Prove by contradiction that there are no integers $m$ and $n$ satisfying $4m+6n=1$.

Mark schemeshow ▾
B1 assume, for contradiction, that integers $m,n$ exist with $4m+6n=1$  M1 $4m+6n=2(2m+3n)$, which is even since $2m+3n$ is an integer  A1 but $1$ is odd, and an even number cannot equal an odd number  A1 contradiction, so no such integers exist ∎.
Q4. 5 marks

Prove by contradiction that $\sqrt{2}$ is irrational. You may use the result that if $n^2$ is even then $n$ is even.

Mark schemeshow ▾
B1 assume, for contradiction, that $\sqrt{2}=\dfrac{a}{b}$ with $a,b$ integers, $b\neq0$, in lowest terms  M1 $a^2=2b^2$, so $a^2$ is even, hence $a$ is even: $a=2c$  M1 $4c^2=2b^2 \Rightarrow b^2=2c^2$, so $b^2$ is even, hence $b$ is even  A1 $a$ and $b$ are both even, contradicting that $\dfrac{a}{b}$ is in lowest terms  A1 so the assumption is false and $\sqrt{2}$ is irrational ∎.
Q5. 5 marks

Prove by contradiction that there is no smallest positive rational number.

Mark schemeshow ▾
B1 assume, for contradiction, that there is a smallest positive rational number $q=\dfrac{a}{b}$ ($a,b$ positive integers)  M1 consider $\dfrac{q}{2}$  A1 $\dfrac{q}{2}=\dfrac{a}{2b}$ is rational and positive  A1 since $q>0$, $\dfrac{q}{2}<q$  A1 so $\dfrac{q}{2}$ is a smaller positive rational, contradicting the assumption; hence no smallest positive rational exists ∎.
Q6. 6 marks

Given that $x$ is irrational and $r$ is rational, prove by contradiction that $r+x$ is irrational.

Mark schemeshow ▾
B1 assume, for contradiction, that $r+x$ is rational  M1 write $r=\dfrac{a}{b}$ and $r+x=\dfrac{c}{d}$ with $a,b,c,d$ integers, $b,d\neq0$  M1 then $x=\dfrac{c}{d}-\dfrac{a}{b}$  A1 $x=\dfrac{bc-ad}{bd}$, a quotient of integers with $bd\neq0$, so $x$ is rational  A1 this contradicts $x$ being irrational  A1 so the assumption is false and $r+x$ is irrational ∎.
Q7. 6 marks

Prove by contradiction that if $n$ is an integer and $n^2$ is divisible by $3$, then $n$ is divisible by $3$.

Mark schemeshow ▾
B1 assume, for contradiction, that $n^2$ is divisible by $3$ but $n$ is not  M1 then $n=3k+1$ or $n=3k+2$ for some integer $k$; square each case  A1 $(3k+1)^2=9k^2+6k+1=3(3k^2+2k)+1$  A1 $(3k+2)^2=9k^2+12k+4=3(3k^2+4k+1)+1$  A1 in both cases $n^2$ leaves remainder $1$ on division by $3$, so $n^2$ is not divisible by $3$ — a contradiction  A1 so the assumption is false and $n$ is divisible by $3$ ∎.
Q8. 7 marks

(a) Prove by contradiction that $\sqrt{3}$ is irrational. You may use the result that if $n^2$ is divisible by $3$ then $n$ is divisible by $3$. (b) Hence prove by contradiction that $5-\sqrt{3}$ is irrational.

Mark schemeshow ▾
(a) B1 assume, for contradiction, that $\sqrt{3}=\dfrac{a}{b}$ with $a,b$ integers, $b\neq0$, in lowest terms  M1 $a^2=3b^2$, so $a^2$ is divisible by $3$, hence $a=3c$  M1 $9c^2=3b^2 \Rightarrow b^2=3c^2$, so $b$ is divisible by $3$  A1 $3$ divides both $a$ and $b$, contradicting lowest terms  A1 so $\sqrt{3}$ is irrational ∎.
(b) M1 assume $5-\sqrt{3}$ is rational; then $\sqrt{3}=5-(5-\sqrt{3})$ is the difference of two rationals, so rational  A1 this contradicts (a), so $5-\sqrt{3}$ is irrational ∎.
Q9. 7 marks

(a) Prove by contradiction that there are no integers $m$ and $n$ satisfying $6m+9n=74$. (b) Given that $a$ and $b$ are real numbers with $a+b>100$, prove by contradiction that at least one of $a$ and $b$ is greater than $50$.

Mark schemeshow ▾
(a) B1 assume, for contradiction, that integers $m,n$ exist with $6m+9n=74$  M1 $6m+9n=3(2m+3n)$, a multiple of $3$  A1 but $74=3\times24+2$ is not a multiple of $3$  A1 contradiction, so no such integers exist ∎.
(b) B1 assume, for contradiction, that $a\le 50$ and $b\le 50$  M1 adding: $a+b\le 100$  A1 this contradicts $a+b>100$, so at least one of $a,b$ is greater than $50$ ∎.
Q10. 8 marks

(a) Prove by contradiction that if $n$ is an integer and $n^3$ is even, then $n$ is even. (b) Hence prove by contradiction that $\sqrt[3]{2}$ is irrational.

Mark schemeshow ▾
(a) B1 assume, for contradiction, that $n^3$ is even but $n$ is odd: $n=2k+1$  M1 $n^3=8k^3+12k^2+6k+1=2(4k^3+6k^2+3k)+1$  A1 which is odd, contradicting $n^3$ even; so $n$ is even ∎.
(b) B1 assume, for contradiction, that $\sqrt[3]{2}=\dfrac{a}{b}$ with $a,b$ integers, $b\neq0$, in lowest terms  M1 $a^3=2b^3$, so $a^3$ is even, hence by (a) $a$ is even: $a=2c$  M1 $8c^3=2b^3 \Rightarrow b^3=4c^3$, so $b^3$ is even, hence $b$ is even  A1 $a$ and $b$ are both even, contradicting lowest terms  A1 so the assumption is false and $\sqrt[3]{2}$ is irrational ∎.