Question

# Find the negation of the following statement: "There exists a negative real number z such that -3<f(z)<=0"

Negative numbers and coordinate plane
Find the negation of the following statement:
"There exists a negative real number z such that -3<f(z)<=0"

2021-02-06

The negation of a statement is the opposite of the statement.
The given statement is "There exists a negative real number z such that $$\displaystyle−{3}{<}{f{{\left({z}\right)}}}\le{0}$$."
The negation of the quantifier "There exist" is "For every".
Also negation of $$\displaystyle−{3}{<}{f{{\left({z}\right)}}}\le{0}$$ is $$\displaystyle{f{{\left({z}\right)}}}\le−{3}{\quad\text{or}\quad}{f{{\left({z}\right)}}}{>}{0}$$.
Therefore, the negation of the statement "There exists a negative real number z such that $$\displaystyle−{3}{<}{f{{\left({z}\right)}}}\le{0}$$" is,
"For every negative real number z, we have $$\displaystyle{f{{\left({z}\right)}}}\le−{3}{\quad\text{or}\quad}{f{{\left({z}\right)}}}{>}{0}$$".