Negations

Number Theory Level 3

Let $S$ be non empty subset of real numbers .

P : There is a rational number $x$ in $S$ such that $x\geq 0$ .

Which of the following is the negation of P ?

There is no rational number

x

S

such that

x <0

Every rational number

x

S

such that

x <0

There is a rational number

x

S

such that

x <0

x

S

and

x<0

implies

x

is not rational

2 solutions

Felipe Knöller
Jan 21, 2017

Since $\overline{(\overline{P})} = P$ , let's say $P$ is the negation of one of the statements. This way is probably easier to find the answer. We can see that the negation of "Every rational number $x$ in $S$ such that $x < 0$ " is "There is AT LEAST one rational number $x$ in $S$ such that $x \geq 0$ ", hence statement $P$ . So, the first statement is the answer.

Prince Loomba
Jan 19, 2017

Hint: The intersection of original and option statement should be phi for negation.

Negations

2 solutions

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