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