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The statement is:
“100 is multiple of 4 and 5”
We know that 100 is a multiple of 4 as well as S. So, p is true statement. Hence, the statement is true i.e. the statement “p” is a valid statement.
“125 is multiple of 5 and 7”
Since 125 is a multiple of 5 but it is not a multiple of 7. So, q is not a true statement i.e. the statement 'q' is not a valid statement.
r: 60 is multiple of 3 or 5
is a compound statement of the following statements:
p: 60 is multiple of 3
q: 60 is multiple of
Suppose q is false. That is, 60 is not a multiple of 5. Clearly p is true.
Thus, if we assume that q is false, then p is true.
Hence, the statement is true i.e. the statement “r” is a valid statement.
Let q and r be the statements given by
q : x and y are odd integers.
r : x + y is an even integer.
Then, the given statement is
if q, then r.
Direct Method: Let q be true. Then,
q is true.
⇒ x and y are odd integers
⇒ x - 2rn + 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 even integer
⇒ r is true.
Thus, q is true ⇒ r is true.
Hence, " if q, then r" is a true statement.
Let r and s be two statements given by
r: xy is an even integer.
s: At least one of x and y is an even integer
Let s be not true. Then,
s is not true Both
Let x = 2n + 1 and y = 2m + 1 for some integers n and m. Then,
⇒ xy -(2n + 1)(2m + 1) for some integers n and m.
⇒ xy 4nm + 2(n + m) + 1 for some integers n and m.
⇒ xy is an odd integer
⇒ xy is not an even integer
⇒ -r is true
Thus, -s is true ⇒ -r is true
Hence, the given statement is true.
Let q and r be the statements given
q: x is a real number such that x3 + x = 0.
r: x is 0.
Then, p: if q, then r.
(i)
q is true
⇒ x is a real number such that x3 + x = 0
⇒ x is a real number such that x (x2 + 1) = 0
⇒ x = 0
Hence, p is true.
(ii)
Method of contrapositive: Let r be not true. 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 ⇒ r is true.
(iii)
Method of contradic6on: If possible, let p be not true. Then,
p is not true
⇒ - p is true
⇒ - (p ⇒ r) is true
⇒ x is a real number such that x3 +x = 0 and x ≠ 0
⇒ x = 0 and x ≠ 0
This a contradiction.
q: If x is an integer and x2 is odd
r: x is an odd integer.
Then, p: "If q, then r."
If possible, let r be false. Then,
r is false
⇒ x is not an odd integer
⇒ x is an even integer
⇒ x - (2n) for some integer
⇒ nx2. 4n2
⇒ x2 is an even integer
⇒ q is false.
Thus, r is false ⇒ q is false.
Hence, p: "if q, then r” is a true statement.
The given statement can be re-written as
“The necessary and sufficient condition that the integer n is even is n2 must be even”
Let p and q be the statements given by
p: the integer n is even.
q. n2 is even.
The given statement is
"p if and only if q"
In order to check its validity, we have to check the validity of the following statements.
(i) 'If p, then q"
(ii) "if q, then p”
Checking the validity of "if p, then q':
The statement "if, then q" is given by:
“if the integer n is even, then n2 is even"
Let us assume that n is even. Then,
n = 2m, where m is an integer
⇒ n2 = (2m)2
⇒ n2 = 4m2
⇒ n2 is an even integer
Thus, n is even ⇒ n2 is even
∴ “if p, then q” is true.
Checking the validity of "if q, then p":
“if n is an integer and n2 is even, then n is even”
Thus, r is false q is false. Hence, p: "if q, then r' is a true statement.
To check the validity of this statements, we will use contrapositive method. So, let n be an odd integer. Then,
⇒ n is odd
⇒ n = 2k +1 for some integer k
⇒ n2 = (2k + 1)2
⇒ n2 = 4k2 - 4k +1
⇒ n2 is not an even integer.
Thus, n is not even ⇒ n2 is not even
∴ "if q, then p" is true.
Hence, "p if and only if q" is true.
Consider a triangle ABC with all angles equal. Then each angle of the triangle is equal to 60°. Hence, ABC is not an obtuse angle triangle.
Therefore the following statement is false.
p: "if all the angles of a triangle are equal, then the triangle is an obtuse angled triangle".
(i) False. Because, no radius of a circle is its chord.
(ii) False. Because, a chord does not have to pass through the centre.
(iii) True. Because a circle is an ellipse that has equal axes.
(iv) True. Because, for any two integers, if x -y is positive then -(x -y) is negative.
(v) False. Because square roots of prime numbers are irrational numbers.
The argument used to check the validity of the given statement is not correct because it does not produce a contradiction.
Chapter 31: Mathematical Reasoning –...