27Prove that if A is not connected then A is not an interval.

{x|x is all real numbers}

The set S contains some real numbers, according to the following three rules.(i) 1/1 is in S.(ii) If a/b is in S, where a/b is written in lowest terms (that is, a and b have highest common factor 1), then b/2ais in S. (iii) If a/b and c/d are in S, where they are written in lowest terms, then a+c/b+d is in S.These rules are exhaustive: if these rules do not imply that a number is in S, then that number is not in S. Can you describe which numbers are in S? For example, by (i), 1/1 is in S. By (ii), since 1/1 is in S, 1/2·1is in S. Since both 1/1 and 1/2 are in S, (iii) tells us 1+1/1+2 is in S. What I have so far: Claim: Set S in contained in interval [½, 1] for a/b where 0<a≤b≤2a The reason is that 1/1 has this form and transformations preserve the property of being in this interval If a≤b≤2a, then b/2a obeys the requirement, since b≤2a≤2b And if a/b and c/d obey the requirement, then so does (a+c)/(b+d), since a+c≤b+d≤2a+2c=2(a+c) However, I feel there is still more to this…

Prove that the product xy of two real numbers x and y is nonnegative if and only if the absolute value | x + y | of their sum is the sum |x | + |y| of their absolute values.

Show that for any real number x and a subset A of R, exactly one of the following holds: (a) x is an interior point of A, (b) x is a boundary point of A or (c) 푥 is an exterior point of A.

Let Q denote the set of rational numbers. Show that card(Q)=x0. What is card(R/Q)?

can someone answer this asap, please

Determine whether the given proposition is true or false, for the universe of all real numbers. Use T for true and F for false. (Væ)(3y)(x + y = 0) (Ja)(Vy)(x² +y = 0) (3æ)(3y)(æ² + y = 0) (Vy)(3)(y = x2)

If a and b are odd positive integers, and the sum of the integers, less than a and greater than b, is 1000; then find a and b.

Let a and b be real numbers. Prove that if 4a+16b= 123, then a and b are not integers