问题1.检查以下陈述的有效性：
(i)p：100是4和5的倍数。
(ii)q：125是5和7的倍数。
(iii)r：60是3或5的倍数。
解决方案：
(i) 100 is completely divisible by 4 and 5 completely without leaving any remainder.Hence, the given statement is true.
(ii) 125 is a multiple of 5 and not divisible perfectly by 7. Hence, the given statement is false.
(iii) The integer 60 is both a multiple of 3 as well as 5. Hence, the given statement is true.
问题2.检查以下陈述是否正确：
(i)p：如果x和y是奇数整数，则x + y是偶数整数。
(ii)q：如果x，y是使得xy为偶数的整数，则x和y中的至少一个是偶数整数。
解决方案：
(i) p: If x and y are odd integers, then x + y is an even integer.
Let us assume that A and B be the components of this statement, which are given by
A: x and y are odd integers.
B: x + y is an even integer
The given statement can be written as :
If A, then B.
Let us assume that A is true. We have, x and y are odd integers.
Let us assume the values to be :
x = 2m+1, y = 2n+1 for some integers m, n
Adding the values , we get
x + y = (2m+1) + (2n+1)
=> x + y = (2m+2n+2)
=> x + y = 2(m+n+1)
x + y is an integer, which is divisible by 2.
Then, B holds true.
Therefore, A is true and B is true.
Hence, if A, then B is a true statement.
(ii) q: if x, y are integer such that xy is even, then at least one of x and y is an even integer.
Let us assume that p and q be the statements given by
p: x and y are integers and xy is an even integer.
q: At least one of x and y is even.
Let p be true, and then xy is an even integer.
So,
xy = 2(n + 1)
Now,
Let x = 2(k + 1)
Since, x is an even integer, xy = 2(k + 1). y is also an even integer.
Now take x = 2(k + 1) and y = 2(m + 1)
xy = 2(k + 1).2(m + 1) = 2.2(k + 1)(m + 1)
So, it is also true.
Hence, the statement is true.
问题3：表明陈述
p：“如果x是一个实数，使得x3 + x = 0，则x为0”是
(i)直接法
(ii)相反的方法
(iii)矛盾的方法
解决方案：
(i) Direct Method:
Let us assume that ‘q’ and ‘r’ be the component statements which are given by
q: x is a real number such that x3 + x=0.
r: x is 0.
We can write this statement as:
If q, then r.
Let q be true. Then, x is a real number such that x3 + x = 0
x is a real number such that x(x2 + 1) = 0
x = 0
r is true
Therefore, q is true and r is true.
Hence, p is true.
(ii) Method of Contrapositive:
Let r be false. Then,
x ≠ 0, x∈R
x(x2+1)≠0, x∈R
q is not true
Thus, -r = -q
Hence, p : q and r is true
(iii) Method of Contradiction:
If possible, let p be false. Then,
P is not true
-p is true
-p (p => r) is true
q and –r is true
x is a real number such that x3+x = 0 and x≠ 0
x = 0 and x ≠ 0
This shows a contradiction.
Hence, p is true.
问题4。通过对立的方法证明以下陈述是正确的
p：“如果x是整数，并且x 2是奇数，则x也是奇数。”
解决方案：
Let us assume that q and r be the component statements of the given compound statement p,
q: x is an integer and x2 is odd.
r: x is an odd integer.
We can rewrite this statement as,
p: if q, then r.
Let us assume r to be false. Then,
If x is not an odd integer, then x is an even integer
x = (2n) for some integer n
Squaring both sides,
x2 = 4n2
x2 is an even integer, since, it is divisible by 2
Therefore, r is false and q is also false.
Hence, p: “ if q, then r” is a true statement.
问题5：证明以下陈述是正确的
“整数n是偶数，当且仅当n 2是偶数时”
解决方案：
Let the component statements,
p: Integer n is even
q: If n2 is even
Let us assume p to be true. Then,
Let n = 2k
On squaring both the sides,
n2 = 4k2
n2 = 2 x 2 x k2
n2 is an even number, since it is divisible by 2.
So, q is also true when p is true.
Therefore, the given statement is true.
问题6.通过给出一个反例，表明以下陈述是不正确的。
p：“如果一个三角形的所有角度都相等，则该三角形是钝角三角形。”
解决方案：
Let us assume any triangle ABC with all angles equal.
Sum of angles of a triangle is 180 degrees, therefore, each angle of the triangle is equal to 60.
So, this triangle ABC is not an obtuse angled triangle.
Hence, the statement “p: If all the angles of a triangle are equal, then the triangle is an obtuse-angled triangle” is False.
问题7.以下哪些陈述是正确的，哪些是错误的?在每种情况下均应给出这样的正当理由
(i)p：圆的每个半径是圆的弦。
(ii)q：圆心一分为二。
(iii)r：圆是椭圆的一种特殊情况。
(iv)s：如果x和y是使得x> y的整数，则– x <– y。
(v)t：√11是一个有理数。
解决方案：
(i) p: Each radius of a circle is a chord of the circle.
A chord of the circle is a line segment joining two end points on its circumference. But radius joins the centre with the end point.
Hence, this statement is False.
(ii) q: The centre of a circle bisect each chord of the circle.
A circle may have many chords. A chord does not necessarily have to pass through the centre of the circle.
Hence, this statement is False.
(iii) r: Circle is a particular case of an ellipse.
If the circle has equal axes, then it is equivalent to an ellipse.
Hence, this statement is true.
(iv) s: If x and y are integers such that x > y, then – x < – y.
For any two integers with x > y , x – y is positive, then –(x-y) is negative.
If we negate the equation, we get
– x < – y
Hence, this statement is true.
(v) t: √11 is a rational number.
Square root of all prime numbers are irrational numbers.
Hence, this statement is False.
问题8.确定用于检查以下语句的有效性的参数是否正确：
p：“如果x 2不合理，则x是有理数。”
该语句为真，因为数×2 =π2是不合理的，因此X =π不合理。
解决方案：
We have the following argument, x2 = π2 is irrational, therefore x = π is irrational.
p: “If x2 is irrational, then x is rational.”
Supposedly, we have an irrational number of the form k, where x = √k, where k is a rational number.
Squaring both sides of the equations, we get,
x2 = k
Now, since k is rational (given), x2 is also rational. This contradicts our statement.
Therefore, the given argument is wrong.