RD Chapter 31 Mathematical Reasoning Ex 31.6 Solutions
Question - 1 : - Check the validity of the following statements:
(i) p: 100 is a multiple of 4 and 5.
(ii) q: 125 is a multiple of 5 and 7.
(iii) r: 60 is a multiple of 3 or 5.
Answer - 1 : -
(i) p: 100 is a multiple of 4 and 5.
We know that 100 is a multiple of 4 as well as 5. So, the given statement is true.
Hence, the statement is true.
(ii) q: 125 is a multiple of 5 and 7
We know that 125 is a multiple of 5 and not a multiple of 7. So, the given statement is false.
Hence, the statement is false.
(iii) r: 60 is a multiple of 3 or 5.
We know that 60 is a multiple of 3 as well as 5. So, the given statement is true.
Hence, the statement is true.
Question - 2 : - Check whether the following statement is true or not:
(i) p: If x and y are odd integers, then x + y is an even integer.
(ii) q : if x, y are integer such that xy is even, then at least one of x and y is an even integer.
Answer - 2 : -
(i) p: If x and y are odd integers, then x + y is an even integer.
Let us assume that ‘p’ and ‘q’ be the statements given by
p: x and y are odd integers.
q: x + y is an even integer
the given statement can be written as :
if p, then q.
Let p be true. Then, x and y are odd integers
x = 2m+1, y = 2n+1 for some integers m, n
x + y = (2m+1) + (2n+1)
x + y = (2m+2n+2)
x + y = 2(m+n+1)
x + y is an integer
q is true.
So, p is true and q is true.
Hence, “if p, then q “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.
Question - 3 : - Show that the statement
p : “If x is a real number such that x3 + x = 0, then x is 0” is true by
(i) Direct method
(ii) method of Contrapositive
(iii) method of contradiction
Answer - 3 : -
(i) DirectMethod:
Let us assume that ‘q’ and ‘r’ be the statements given by
q: x is a real number such that x3 + x=0.
r: x is 0.
The given statement can be written 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
Thus, q is true
Therefore, q is true and r is true.
Hence, p is true.
(ii) Methodof Contrapositive:
Let r be false. Then,
R is not true
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) Methodof 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 = 0and x≠ 0
x =0 and x≠0
This is a contradiction.
Hence, p is true.
Question - 4 : - Show that the following statement is true by the method of the contrapositive
p: “If x is an integer and x2 is odd, then x is also odd.”
Answer - 4 : -
Let us assume that ‘q’ and ‘r’ be the statements given
q: x is an integer and x2 is odd.
r: x is an odd integer.
The given statement can be written as:
p: if q, then r.
Let r be false. Then,
x is not an odd integer, then x is an even integer
x = (2n) for some integer n
x2 = 4n2
x2 is an even integer
Thus, q is False
Therefore, r is false and q is false
Hence, p: “ if q, then r” is a true statement.
Question - 5 : - Show that thefollowing statement is true
“The integer n is even if and only if n2 iseven”
Answer - 5 : -
Let the statements,
p: Integer n is even
q: If n2 is even
Let p be true. Then,
Let n = 2k
Squaring both the sides, we get,
n2 = 4k2
n2 = 2.2k2
n2 is an even number.
So, q is true when p is true.
Hence, the given statement is true.
Question - 6 : - By giving a counterexample, show that the following statement is not true.
p: “If all the angles of a triangle are equal, thenthe triangle is an obtuse angled triangle.”
Answer - 6 : -
Let us consider a triangle ABC with all angles equal.
Then, each angle of the triangle is equal to 60.
So, ABC is not an obtuse angle triangle.
Hence, the statement “p: If all the angles of a triangle areequal, then the triangle is an obtuse angled triangle” is False.
Question - 7 : - Which of thefollowing statements are true and which are false? In each case give a validreason for saying so
(i) p: Each radius of a circle is a chord of thecircle.
(ii) q: The centre of a circle bisect each chord ofthe circle.
(iii) r: Circle is a particular case of an ellipse.
(iv) s: If x and y are integers such that x > y,then – x < – y.
(v) t: √11 is a rational number.
Answer - 7 : -
(i) p: Eachradius of a circle is a chord of the circle.
The Radius of the circle is not it chord.
Hence, this statement is False.
(ii) q:The centre of a circle bisect each chord of the circle.
A chord does not have to pass through the center.
Hence, this statement is False.
(iii) r:Circle is a particular case of an ellipse.
A circle can be an ellipse in a particular case when the circlehas equal axes.
Hence, this statement is true.
(iv) s:If x and y are integers such that x > y, then – x < – y.
For any two integers, if x – y id positive then –(x-y) isnegative.
Hence, this statement is true.
(v) t:√11 is a rational number.
Square root of prime numbers is irrational numbers.
Hence, this statement is False.
Question - 8 : - Determine whetherthe argument used to check the validity of the following statement is correct:
p: “If x2 is irrational, then x isrational.”
The statement is true because the number x2 =π2 is irrational, therefore x = π is irrational.
Answer - 8 : -
Argument Used: x2 = π2 isirrational, therefore x = π is irrational.
p: “If x2 is irrational, then x is rational.”
Let us take an irrational number given by x = √k, where k is arational number.
Squaring both sides, we get,
x2 = k
x2 is a rational number and contradicts ourstatement.
Hence, the given argument is wrong.