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IAL 2021 Oct Q10

A Level / Edexcel / P4

IAL 2021 Oct Paper · Question 10

题目

Problem

(a) A student’s attempt to answer the question

“Prove by contradiction that if n3n^3 is even, then nn is even”

is shown below. Line 5 of the proof is missing.

Complete this proof by filling in line 5.

(1)

(b) Hence, prove by contradiction that 23\sqrt[3]{2} is irrational.

(5)
题目中文翻译

(a) 某位学生尝试回答下列问题:

“用反证法证明:如果 n3n^3 是偶数,那么 nn 是偶数”

其证明过程如下所示,其中第 5 行缺失。

补全这个证明,填入第 5 行。

(b) 进而用反证法证明 23\sqrt[3]{2} 是无理数。

解答

(a)

解法一

思路

展开

把展开式中除常数 11 外的各项提出公因数 22,便可把 n3n^3 写成“偶数加 11”的形式,从而说明它是奇数并得到矛盾。

答题过程

展开

The missing line is

n3=2(4p3+6p2+3p)+1,which is odd.\boxed{ \begin{gathered} n^3=2\big(4p^3+6p^2+3p\big)+1,\\ \text{which is odd.} \end{gathered} }

(b)

解法一

思路

展开

承接 (a),反设 23\sqrt[3]{2} 是有理数,并写成最简分数 p/qp/q。立方后先证明 p3p^3 为偶数,因此 pp 为偶数;代入 p=2mp=2m 后又能证明 q3q^3 为偶数,因此 qq 也为偶数。这与 p/qp/q 已是最简分数矛盾。

答题过程

展开

Assume, for a contradiction, that 23\sqrt[3]{2} is rational. Then there exist integers pp and qq, with q0q\neq0, such that

23=pq,\sqrt[3]{2}=\frac{p}{q},

where pp and qq have no common factor.

Cubing both sides gives

2=p3q3,2=\frac{p^3}{q^3},

so

p3=2q3.p^3=2q^3.

Thus p3p^3 is even. By the result in part (a), pp is even, so p=2mp=2m for some integer mm.

Substituting p=2mp=2m into p3=2q3p^3=2q^3 gives

(2m)3=2q3.(2m)^3=2q^3.

Hence,

8m3=2q3q3=4m3.8m^3=2q^3 \quad\Longrightarrow\quad q^3=4m^3.

Therefore q3q^3 is even, so part (a) implies that qq is also even.

Both pp and qq are therefore divisible by 22, contradicting the assumption that p/qp/q is in its simplest form. Hence,

23 is irrational.\boxed{\sqrt[3]{2}\text{ is irrational}.}