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
ANSWERED
asked 2021-02-05
Find the negation of the following statement:
"There exists a negative real number z such that -3<f(z)<=0"

Answers (1)

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 \(\displaystyle{f{{\left({z}\right)}}}\le−{3}{\quad\text{or}\quad}{f{{\left({z}\right)}}}{>}{0}\)".

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