Qus1
(A)-
1 2 3 5 4 6
F F T F F T
F T F T T T
T F T F T T
T T F F F T
Tautology of statement
"Changed in the table so as not to get similarity"
(B)-
1
2
3
4
5
Tautology
Qus−2: It is clear that, P(x) is true for all values as, P(1) is True, P(2) is true. Thus, the truth value is (((True)))).
It is clear that P(0.5) is True, but P(x) is False for x = -1 (take x = -1 as a counterexample). Thus, the truth value of is ((( False))). It is clear that ,P(x) is False for all real values except, x=1 (take x = 1 as a counterexample) where, P (1) is true. Thus, the truth value of is (((True))).
It is clear that, P(x) is False for every value as ,x =1 (take x = 1 as a counterexample). Thus, the truth
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Then a | (b + c) 2 | 4 but a ∤b because (2∤3 )and( a∤c )because( 2∤1).
b
(T) True:
Assume that is (a|c) and (b|c).
Then, from the placement of divisibility, it follows that there are integers s and t with c=as and c=bt. Hence, c2=as bt = ab st. Therefore, ab divides c2.
If we let a=2 b=3, and c=6. Then( a|c) because (2|6) and (b|c) because (3|6) then (ab)|c2 because (6|36) .
c
False:
Assume that a|b and c|d. Then, from the definition of divisibility, it follows that there are integers s and t with b=as and d=ct. Hence, b+d=as+ct . Therefore, ac not divides b+d.
IF we Let a=2, b=4, c=5 and d=10. Then a|b because 2|4 and c | d because 5|10 then
(ac) ∤ (b+d) because 10∤14.
d
True:
Assume that a|b and b|c. Then, from the definition of divisibility, it follows that there are integers s and t with b=as, c=bt and d=cx. Hence, d=btx=astx=a stx. Therefore, a divides d.
Let a=2, b=4, c=8 and d=16. Then a|b because 2|4, b|c because 4|8 and c|d because 8|16 then a|d because 2|16.
e
False:
Assume that a|b and c|d Then, from the definition of divisibility, it follows that there are integers s and t with b=as and d=ct. Hence, b+d=as+ct. Therefore, a+c not divides
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(b) b-
(T)True:
Where 2 | 6, we can see immediately that 6 is not a minimal element of A.
(c) c-