Solution to Let Q(x)Q(x) be the statement “x + 1 > 2xx+1>2x ”. If the domain consists … - Sikademy
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Let Q(x)Q(x) be the statement “x + 1 > 2xx+1>2x ”. If the domain consists of all integers, let us find the following truth values. a) Since 1>0,1>0, we get that 𝑄(0)Q(0) is true. b) Taking into account that it is not true that 0>-2,0>−2, we get that 𝑄(-1)Q(−1) is false. c) Taking into account that it is not true that 2>2,2>2, we get that 𝑄(1)Q(1) is false. d) Since 1>0,1>0, we get that 𝑄(0)Q(0) is true, and hence ∃𝑥𝑄(𝑥)∃xQ(x) is true. e) Taking into account that it is not true that 0>-2,0>−2, we get that 𝑄(-1)Q(−1) is false, and hence ∀𝑥𝑄(𝑥)∀xQ(x) is false. f) Since it is not true that 2>2,2>2, we get that 𝑄(1)Q(1) is false. Therefore, \neg 𝑄(1)¬Q(1) is true, and thus ∃𝑥¬𝑄(𝑥)∃x¬Q(x) is true. g) Since 1>0,1>0, we get that 𝑄(0)Q(0) is true. We conclude that \neg 𝑄(0)¬Q(0) is false, and hence ∀𝑥¬𝑄(𝑥)∀x¬Q(x) is false.

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Let Q(x) be the statement “x + 1 > 2x ”. If the domain consists of all integers, let us find the following truth values.


a) Since 1>0, we get that 𝑄(0) is true.


b) Taking into account that it is not true that 0>-2, we get that 𝑄(-1) is false.


c) Taking into account that it is not true that 2>2, we get that 𝑄(1) is false.


d) Since 1>0, we get that 𝑄(0) is true, and hence ∃𝑥𝑄(𝑥) is true.


e) Taking into account that it is not true that 0>-2, we get that 𝑄(-1) is false, and hence ∀𝑥𝑄(𝑥) is false.


f) Since it is not true that 2>2, we get that 𝑄(1) is false. Therefore, \neg 𝑄(1) is true, and thus ∃𝑥¬𝑄(𝑥) is true.


g) Since 1>0, we get that 𝑄(0) is true. We conclude that \neg 𝑄(0) is false, and hence ∀𝑥¬𝑄(𝑥) is false.

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Question ID: mtid-5-stid-8-sqid-1239-qpid-977