Negation of $\exists a \in \Bbb R$ such that $a >0$ and $\forall x \in \Bbb R$; if $x \ge a$, then $2^{-x} < \frac{1}{50}$.

Question

To negate the following statement:

$\exists a \in \Bbb R$ such that $a >0$ and $\forall x \in \Bbb R$; if $x \ge a$, then $2^{-x} < \frac{1}{50}$.

Note that for negation we do the following:

"A and B" changes to "not A or not B".

"if A, then B" changes to "A and not B".

"For all $x, A(x)$" changes to "There exist $x$ such that not $A(x)$".

"There exists $x$ such that $A(x)$" changes to "For every $x$, not $A(x)$".

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So my result after negating the given statement is:

$\forall a \in \Bbb R; a \le 0$ or $\exists x \in \Bbb R$, $x \ge a$ and $2^{-x} \ge \frac{1}{50}$.

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Is the negation of the given statement correct?

Thank You.

Answer 1

$\forall a \in \mathbb{R} \text{ with } a > 0 \exists x \in \mathbb{R}: x \geq a \text{ and } 2^{-x} \geq 1/50$.

Answer 2

First write the statement formally (changing restricted quantifiers to implications):

$$\exists a \in \Bbb R: (a >0) \land \left( \forall x \in \Bbb R: (x \ge a) \to (2^{-x} < \frac{1}{50})\right)$$

The negation of the implication is

$$(x \ge a) \land (2^{-x} \ge \frac{1}{50})$$ as the only way to make an implication false is to make the antecedent true and the consequent false.

So the negation of the statement between braces, with the quantor, is

$$\exists x \in \Bbb R: (x \ge a) \land (2^{-x} \ge \frac{1}{50})$$

So the negation of the $\land$-statement is the $\lor$ of the negations so

$$(a \le 0) \lor \left( \exists x \in \Bbb R: (x \ge a) \land (2^{-x} \ge \frac{1}{50}) \right)$$

and taking care of the outer quantifier:

$$\forall a \in \Bbb R: (a \le 0) \lor \left( \exists x \in \Bbb R: (x \ge a) \land (2^{-x} \ge \frac{1}{50}) \right)$$

which is the negation of the original statement.

This seems to match yours.

If you think about it, the statement is trivial for $a \le 0$, so we see this last statement is equivalent to

$$\forall a >0: \exists x \in \Bbb R: (x \ge a) \land (2^{-x} \ge \frac{1}{50}) $$