Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Exercise 1.5 [Page 12]
Use quantifiers to convert the following open sentences defined on N (Natural Numbers), into a true statement.
\(x^2 + 3x - 10 = 0\)
To make the open sentence \(x^2 + 3x - 10 = 0\) a true statement on N, we need to find if there exists at least one natural number \(x\) that satisfies this equation.
The equation is a quadratic: \(x^2 + 3x - 10 = 0\). Factoring it, we get \((x+5)(x-2) = 0\). The solutions are \(x = -5\) or \(x = 2\).
Since \(x\) must be a natural number (\(x \in N = \{1, 2, 3, \dots\}\)), \(x = 2\) is a valid solution. \(x = -5\) is not a natural number.
Thus, we can use the existential quantifier:
\(\exists x \in N\), such that \(x^2 + 3x - 10 = 0\)
This is a true statement, because \(x = 2 \in N\) satisfies it: \(2^2 + 3(2) - 10 = 4 + 6 - 10 = 0\).
Use quantifiers to convert the following open sentences defined on N, into a true statement.
\(3x - 4 < 9\)
The open sentence is \(3x - 4 < 9\), defined on N.
Simplifying the inequality: \[3x - 4 < 9\] \[3x < 9 + 4\] \[3x < 13\] \[x < \frac{13}{3}\] \[x < 4.33\dots\]
We need to find natural numbers \(x \in N = \{1, 2, 3, \dots\}\) such that \(x < 4.33\dots\). The natural numbers satisfying this are \(x = 1, 2, 3, 4\).
Since there exist such natural numbers, we use the existential quantifier:
\(\exists x \in N\), such that \(3x - 4 < 9\)
This is a true statement. For example, if \(x = 1 \in N\), \(3(1) - 4 = -1 < 9\). If \(x = 4 \in N\), \(3(4) - 4 = 8 < 9\).
Use quantifiers to convert the following open sentences defined on N, into a true statement.
\(n^2 \ge 1\)
The open sentence is \(n^2 \ge 1\), defined on N.
Let's check for natural numbers \(n \in N = \{1, 2, 3, \dots\}\):
- If \(n=1\), \(1^2 = 1\), which is \(\ge 1\). (True)
- If \(n=2\), \(2^2 = 4\), which is \(\ge 1\). (True)
- If \(n=3\), \(3^2 = 9\), which is \(\ge 1\). (True)
For any natural number \(n\), \(n \ge 1\). Squaring both sides (since both are positive), we get \(n^2 \ge 1^2\), so \(n^2 \ge 1\). This holds for all natural numbers.
Thus, we use the universal quantifier:
\(\forall n \in N, n^2 \ge 1\)
This is a true statement, as it holds for every natural number \(n\).
Use quantifiers to convert the following open sentences defined on N, into a true statement.
\(2n - 1 = 5\)
The open sentence is \(2n - 1 = 5\), defined on N.
Solving for \(n\): \[2n - 1 = 5\] \[2n = 5 + 1\] \[2n = 6\] \[n = 3\]
Since \(n = 3\) is a natural number (\(3 \in N\)), there exists a natural number that satisfies the equation.
Thus, we use the existential quantifier:
\(\exists n \in N\), such that \(2n - 1 = 5\)
This is a true statement because \(n = 3 \in N\) satisfies the equation: \(2(3) - 1 = 6 - 1 = 5\).
Use quantifiers to convert the following open sentences defined on N, into a true statement.
\(y + 4 > 6\)
The open sentence is \(y + 4 > 6\), defined on N.
Simplifying the inequality: \[y + 4 > 6\] \[y > 6 - 4\] \[y > 2\]
We need to find natural numbers \(y \in N = \{1, 2, 3, \dots\}\) such that \(y > 2\). Examples of such natural numbers are \(y = 3, 4, 5, \dots\).
Since there exist such natural numbers, we use the existential quantifier:
\(\exists y \in N\), such that \(y + 4 > 6\)
This is a true statement. For example, if \(y = 3 \in N\), \(3 + 4 = 7 > 6\).
Use quantifiers to convert the following open sentences defined on N, into a true statement.
\(3y - 2 \le 9\)
The open sentence is \(3y - 2 \le 9\), defined on N.
Simplifying the inequality: \[3y - 2 \le 9\] \[3y \le 9 + 2\] \[3y \le 11\] \[y \le \frac{11}{3}\] \[y \le 3.66\dots\]
We need to find natural numbers \(y \in N = \{1, 2, 3, \dots\}\) such that \(y \le 3.66\dots\). The natural numbers satisfying this are \(y = 1, 2, 3\).
Since there exist such natural numbers, we use the existential quantifier:
\(\exists y \in N\), such that \(3y - 2 \le 9\)
This is a true statement. For example, if \(y = 1 \in N\), \(3(1) - 2 = 1 \le 9\). If \(y = 3 \in N\), \(3(3) - 2 = 7 \le 9\).
If \(B = \{2, 3, 5, 6, 7\}\), determine the truth value of \(\forall x \in B\) such that \(x\) is a prime number.
The given set is \(B = \{2, 3, 5, 6, 7\}\).
The statement is "\(\forall x \in B\), \(x\) is a prime number". This means every element in B must be a prime number for the statement to be true.
Let's check each element of B:
- Is 2 prime? Yes (divisors are 1 and 2).
- Is 3 prime? Yes (divisors are 1 and 3).
- Is 5 prime? Yes (divisors are 1 and 5).
- Is 6 prime? No (divisors are 1, 2, 3, 6). Since 6 is divisible by 2 and 3 (other than 1 and itself), it is not prime.
- Is 7 prime? Yes (divisors are 1 and 7).
Since the element \(x = 6 \in B\) is not a prime number, the statement that "all \(x \in B\) are prime" is false.
Therefore, the truth value of the given statement is F.
If \(B = \{2, 3, 5, 6, 7\}\), determine the truth value of \(\exists n \in B\), such that \(n + 6 > 12\).
The given set is \(B = \{2, 3, 5, 6, 7\}\).
The statement is "\(\exists n \in B\), such that \(n + 6 > 12\)". This means there must be at least one element in B that satisfies the condition \(n + 6 > 12\).
The condition is \(n + 6 > 12\), which simplifies to \(n > 12 - 6\), so \(n > 6\).
Let's check elements of B to see if any satisfy \(n > 6\):
- For \(n = 2\): Is \(2 > 6\)? No.
- For \(n = 3\): Is \(3 > 6\)? No.
- For \(n = 5\): Is \(5 > 6\)? No.
- For \(n = 6\): Is \(6 > 6\)? No.
- For \(n = 7\): Is \(7 > 6\)? Yes.
Since \(n = 7 \in B\) satisfies the condition (\(7 + 6 = 13\), and \(13 > 12\)), the statement "\(\exists n \in B\), such that \(n + 6 > 12\)" is true.
Therefore, its truth value is T.
If \(B = \{2, 3, 5, 6, 7\}\), determine the truth value of \(\exists n \in B\), such that \(2n + 2 < 4\).
The given set is \(B = \{2, 3, 5, 6, 7\}\).
The statement is "\(\exists n \in B\), such that \(2n + 2 < 4\)". This means there must be at least one element in B that satisfies the condition \(2n + 2 < 4\).
The condition is \(2n + 2 < 4\), which simplifies to \(2n < 4 - 2\), so \(2n < 2\), or \(n < 1\).
Let's check elements of B to see if any satisfy \(n < 1\):
- For \(n = 2\): Is \(2 < 1\)? No.
- For \(n = 3\): Is \(3 < 1\)? No.
- For \(n = 5\): Is \(5 < 1\)? No.
- For \(n = 6\): Is \(6 < 1\)? No.
- For \(n = 7\): Is \(7 < 1\)? No.
There is no element \(n\) in B such that \(n < 1\).
Therefore, the statement "\(\exists n \in B\), such that \(2n + 2 < 4\)" is false.
Its truth value is F.
If \(B = \{2, 3, 5, 6, 7\}\), determine the truth value of \(\forall y \in B\), such that \(y^2\) is negative.
The given set is \(B = \{2, 3, 5, 6, 7\}\).
The statement is "\(\forall y \in B\), such that \(y^2\) is negative". This means \(y^2 < 0\) for every element \(y\) in B.
The elements of B are all positive integers. The square of any non-zero real number is always positive. The square of zero is zero. No real number squared results in a negative number.
Let's check for any \(y \in B\):
- For \(y = 2\), \(y^2 = 2^2 = 4\). Is \(4 < 0\)? No.
Since the condition \(y^2 < 0\) is false for \(y=2\) (and in fact, for all elements in B, as their squares will be \(4, 9, 25, 36, 49\) respectively, all of which are positive), the universal statement "\(\forall y \in B\), such that \(y^2\) is negative" is false.
Therefore, its truth value is F.
If \(B = \{2, 3, 5, 6, 7\}\), determine the truth value of \(\forall y \in B\), such that \((y - 5) \in N\).
Note: N represents the set of Natural Numbers, typically \(\{1, 2, 3, \dots\}\). We will assume this definition.
The given set is \(B = \{2, 3, 5, 6, 7\}\).
The statement is "\(\forall y \in B\), such that \((y - 5) \in N\)". This means for every element \(y\) in B, the result of \(y-5\) must be a natural number.
Let's check each \(y \in B\):
- For \(y = 2\): \(y - 5 = 2 - 5 = -3\). Is \(-3 \in N\)? No, because -3 is not a natural number.
Since the condition \((y - 5) \in N\) is false for \(y = 2\), the universal statement "\(\forall y \in B\), such that \((y - 5) \in N\)" is immediately false. We don't need to check further for a universal statement if one counterexample is found.
(For completeness, let's check other values:
For \(y = 3\), \(y - 5 = -2\). \(-2 \notin N\).
For \(y = 5\), \(y - 5 = 0\). \(0 \notin N\) (assuming N starts from 1).
For \(y = 6\), \(y - 5 = 1\). \(1 \in N\). (True for this case)
For \(y = 7\), \(y - 5 = 2\). \(2 \in N\). (True for this case))
However, because the statement must hold for ALL \(y \in B\), and it fails for \(y=2\) (and \(y=3, y=5\)), the overall statement is false.
Therefore, its truth value is F.
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