Question #266489

Write down the negation of the following statements and determine the truth value of the negation:

a) āˆ€š‘„ ∈ š‘¹, š‘„ 2 + 1 ≄ 2š‘„

b) āˆ€š‘„ ∈ š‘¹, (š‘¦ ≠→ (š‘¦ + 1)/š‘¦ < 1

c) āˆƒš‘§ ∈ š’, (š‘§ š‘–š‘  š‘œš‘‘š‘‘) ∨ (š‘§ š‘–š‘  š‘’š‘£š‘’š‘›)

d) āˆƒš‘› ∈ š‘µ, (š‘› š‘–š‘  š‘’š‘£š‘’š‘›) ∧ (āˆšš‘› š‘–š‘  š‘š‘Ÿš‘–š‘šš‘’)


Expert's answer

a)∃x∈R,x2+1<2x\exist x \isin R, x^2 +1<2x. False, because least value of x2āˆ’2x+1=(xāˆ’1)2x^2 -2x +1=(x-1)^2 is 0 in case x=1.

b)∃y∈R,y≠0∧(y+1)/y≄1\exist y \isin R, y ≠ 0 ∧ (y+1)/y ≄ 1. True, because for example with any y>0y>0 , 1+1/y≄11+1/y ≄1.

c)āˆ€z∈Z,\forall z \isin Z, (z is not odd)∧(z is not even)(z \space is \space not \space odd) ∧ (z \space is \space not \space even). False, because if we choose any odd or even integer, and statement become false.

d)āˆ€n∈N,(n is not even)∨(√n is not prime)\forall n \isin N, (n \space is \space not \space even) ∨ (√n \space is \space not \space prime). False, because for example n=4, in this case n is even and √n=√4=2 is prime.


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