Question #132111
(1) Let Q(x, y) be the statement "x+y=xy." If the domain for both variables consists of all integers, what are the truth values?
(a)Q(1,1)
(b)Q(2,0)
(c)∀yQ(1, y)
(d)∃xQ(x,2)
(e)∃x∃yQ(x, y)
(f)∀x∃yQ(x, y)
(g)∃y∀xQ(x, y)
(h)∀y∃xQ(x, y)
(i)∀x∀yQ(x, y)
1
Expert's answer
2020-09-09T19:35:53-0400

a. Q(1,1)


x+y=1+1=2x+y=1+1=2xy=(1)(1)=1xy=(1) (1)=1

Since x+y is not equal to xy, truth value of Q(x, y) is FALSE

Answer: FALSE


b. Q(2,0)


x+y=2+0=2x+y=2+0=2xy=20=0xy=2*0=0

Since x+y is not equal to xy, the truth value of Q(x, y) is FALSE

Answer: FALSE


c. yQ(1,y)\forall yQ(1,y)

By contradiction;

let y be 1:


x+y=1+1=2x+y=1+1=2xy=11=1xy=1*1=1

Since x+y is not equal to xy, the truth value of Q(x, y) is FALSE

Answer: FALSE


d. xQ(x,2)\exist xQ(x, 2)

let x be 2:


x+y=2+2=4x+y=2+2=4xy=22=4xy=2*2=4

Since x+y=xy, the truth value of Q(x, y) is TRUE

Answer : TRUE


e. xyQ(x,y)\exist x\exists yQ(x, y)

let x=2 and y=2


x+y=2+2=4x+y=2+2=4xy=22=4xy=2*2=4

Since x+y=xy, the truth value of Q(x, y) is TRUE

Answer : TRUE


f. xyQ(x,y)\forall x \exists yQ(x, y)

Let x=1 and y=1:


x+y=1+1=2x+y=1+1=2xy=11=1xy=1*1=1

Since x+y is not equal to xy, then the truth value of Q(x, y) is FALSE

Answer : FALSE


g. yxQ(x,y)\exist y \forall xQ(x, y)

Let x=1 and y=1:


x+y=1+1=2x+y=1+1=2xy=11=1xy=1* 1=1


Since x+y is not equal to xy, then the truth value of Q(x, y) is FALSE


Answer : FALSE


h. yxQ(x,y)\forall y \exist xQ(x, y)

Let x=1, y=1:


x+y=1+1=2x+y=1+1=2xy=11=1xy=1* 1=1

Since x+y is not equal to xy, then the truth value of Q(x, y) is FALSE


Answer : FALSE


i. xyQ(x,y)\forall x\forall yQ(x, y)

Let x=1, y=1:


x+y=1+1=2x+y=1+1=2xy=11=1xy=1*1=1

Since x+y is not equal to xy, then the truth value of Q(x, y) is FALSE


Answer : FALSE

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Comments

Assignment Expert
14.09.20, 21:08

Dear Promise Omiponle. You changed conditions of the question considerably. Please use a panel for submitting new questions.

Promise Omiponle
11.09.20, 19:45

Hello, please can this question #132111 in Discrete Mathematics be solved again? I made a huge mistake in the question. Q(x,y) is supposed to be "x+y=x-y". Please can it be resolved and posted today?

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