第 11 课 NCERT 解决方案 - 第 14 章数学推理 - 练习 14.5
问题 1：证明语句
p: '如果 x 是一个实数，使得 x 3 + 4x = 0，那么 x 是 0' 为真
(一) 直接法
解决方案：
p: ‘If x is a real number such that x3 + 4x = 0, then x is 0’.
Let q: x is a real number such that x3 + 4x = 0
r: x is 0.
(i) To show that statement p is true, we assume that q, is true and then show that r is true.
Therefore, let statement q be true.
x3 + 4x = 0
x(x2 + 4) = 0
x = 0 or x2 + 4 = 0
However, since x is real, it is 0.
Thus, statement r is true.
Therefore, the given statement is true.
(ii) 矛盾法
解决方案：
To show statement p to be true by contradiction, we assume that p is not true.
Let x be a real number such that x3 + 4x = 0 and let x is not 0.
Therefore, x3 + 4x = 0
x (x2 + 4) = 0
x = 0 or x2 + 4 = 0
x = 0 or x2 = –4
However, x is real. Therefore, x = 0, which is a contradiction since we have assumed that x is not 0.
(iii) 反证法
解决方案：
To prove statement p to be true by contrapositive method, we assume that r is false and prove that q must be false.
Here, r is false implies that it is required to consider the negation of statement r. This obtains the following statement.
~ r: x is not 0
I can be seen that (x2 + 4) will always be positive
x = 0 implies that the product of any positive real number with x is not zero.
Let us consider the product of x with (x2 + 4)
⸫ x(x2 + 4) = 0
x3 + 4x = 0
This shows that statement q is not true.
Thus, it has been proved that
~ r
⇒ ~ q
Therefore, the given statement p is true.
问题 2：通过给出一个反例来证明“对于任何实数 a 和 b，a 2 = b 2意味着 a = b”这句话是不正确的。
解决方案：
The given statement can be written in the form of ‘if-then’ as follows.
If a and b are real numbers such that a2 = b2 , then a = b.
Let p: a and b are real numbers such that a2 = b2.
q: a = b
The given statement has to be proved false. For this purpose, it has to be proved that if p, then ~ q.
To show this, two real numbers, a and b, with a2 = b2 are required such that a ≠ b.
Let a = 1 and b = –1
a2 = (1)2 and b2 = (–1)2 = 1
a2 = b2
However, a = b
Thus, it can be concluded that the given statement is false.
问题3：用反证法证明下列陈述为真。 p：如果x是整数并且x 2是偶数，那么x也是偶数。
解决方案：
p: If x is an integer and x2 is even, then x is also even.
Let q: x is an integer and x2 is even.
r: x is even.
To prove that p is true by contrapositive method, we assume that r is false, and prove that q is also false.
Let x is not even.
To prove that q is false, it has to be proved that x is not an integer or x2 is not even.
x is not even implies that x2 is also not even.
Therefore, statement q is false.
Thus, the given statement p is true.
问题 4：举一个反例，证明下列陈述不正确。
(i) p：如果一个三角形的所有角都相等，那么这个三角形就是一个钝角三角形。
解决方案：
The given statement is of the form ‘if q then r’.
q: All the angles of a triangle are equal.
r: The triangle is an obtuse-angled triangle.
The given statement p has to be proved false. For this purpose, it has to be proved that if q, then ~ r.
To show this, angles of a triangle are required such that none of them is an obtuse angle.
It is known that the sum of all angles of a triangle is 180°. Therefore, if all the three angles
are equal, then each of them is of measure 60°, which is not an obtuse angle.
In an equilateral triangle, the measure of all angles is equal. However, the triangle is not an obtuse-angled triangle.
Thus, it can be concluded that the given statement p is false.
( ii) q：方程 x 2 – 1 = 0 在 0 和 2 之间没有根。
解决方案：
The given statement is as follows.
q: The equation x2 – 1 = 0 does not have a root lying between 0 and 2.
This statement has to be proved false. To show this, a counter example is required.
Consider x2– 1 = 0
x2 = 1
x = ±1
One root of the equation x2 – 1 = 0, i.e. the root x = 1, lies between 0 and 2.
Thus, the given statement is false.
问题5：下列哪些说法是正确的，哪些是错误的？在每种情况下都给出这样说的正当理由。
(i) p：圆的每个半径都是圆的弦。
解决方案：
The given statement p is false.
According to the definition of chord, it should intersect the circle at two distinct points.
(ii) q：圆心平分圆的每个弦。
解决方案：
The given statement q is false.
If the chord is not the diameter of the circle, then the centre will not bisect that chord.
In other words, the centre of a circle only bisects the diameter, which is the chord of the
circle.
(iii) r：圆是椭圆的特例。
解决方案：
The equation of an ellipse is,
X2/a2 + Y2/b2 = 1
If we put a = b = 1, then we obtain
x2 + y2 = 1,which is an equation of a circle
Therefore, circle is a particular case of an eclipse.
Thus, statement r is true.
(iv) s：如果 x 和 y 是整数，使得 x > y，则 – x < – y。
解决方案：
x > y
⇒ – x < – y (By a rule of inequality)
Thus, the given statement s is true.
(v) t：√11 是一个有理数。
解决方案：
11 is a prime number, and we know that the square root of any prime number is an irrational number.
Therefore,
√11 is an irrational number.
Thus, the given statement t is false.