Let A and B be sets of real numbers, let f be a function from R to R, and let P be the set?

Question

of positive real numbers. Without using words of negation, for each statement write a sentence that expresses its negation.

a)For all xЄA, there is a bЄB such that b>x
b)There is an xЄA such that, for all bЄB, b>x
c)For all x, yЄR, f(x) = f(y) => x=y
d)For all bЄR, there is an xЄR such that f(x)=b
e)For all x, yЄR and all εЄP, there is a δЄP such that |x-y|<δ implies |f(x)-f(y)|<ε
d)For all εЄP, there is a δЄP such that, for all x, yЄR, |x-y|<δ implies |f(x)-f(y)|<ε

Any help on this would be nice!

Answer 1

Solving this means having knowledge about negation in quantifier logic. The negation of "for all" is "there exists", and the negation of "there exists" is "for all".

a) For all x in A, there is a b in B such that b > x

Negating this piece by piece,

There exists x in A, such that for all b in B, b <= x.

b) There exists an x in A such that for all b in B, b > x.

Negating piece by piece,

For all x in A, there exists a b in B, such that b <= x.

Answer 2

a. For some x there exists b such that b <= x

b. Step 1. For all x the rest is NOT true.
Step 2. For all x, there exists b such that the rest is NOT true.
Step 3. For all x, there exists b such that b <= x.

d. Step 1. There exists b such that the rest is not true.
Step 2. There exists b such that for all x the rest is not true.

And so on.

In most cases, an important point is that "There exists an X such that ABC is true" is the negation of "For all X, ABC is false" and vice versa.

Answer 3

The closure of R Q is R: all reduce factors of R Q are authentic numbers (what else ought to they be?) no longer merely R Q is dense in R, yet also Q is dense in R it truly is a lot extra spectacular in case you imagine that Q has the cardinality of N, it truly is Q is a countable set! besides it is really exciting to locate 2 disjoint subsets both dense