Part a
∃x(x2=2)Here x=∓2 so satisfy\exist x (x^2 =2)\\ Here \space x= ∓ \sqrt{2} \space \space so \space satisfy∃x(x2=2)Here x=∓2 so satisfy. Ture
Part b
∃x(x2=1)\exist x (x^2 =1)\\∃x(x2=1)
Since for no real number, the square can be negative. So it is false.
Part c
∀x(x2+2∗)\forall x (x^2+2*)∀x(x2+2∗)
* These symbols are not proper of the reformulate up to the best of my knowledge ∀x(x2∧x≠1)\forall x (x^2 \land x \not= 1)∀x(x2∧x=1) False
Part d
∀x(x2+x)\forall x (x^2+x)∀x(x2+x)
Since for every real number x2+x has a solution. True
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