For $X,Y \subseteq \Bbb R$ define $X+Y : \{x + y \mid x \in X, y \in Y\}$ examples where $X + Y \neq \Bbb R$ are

(A.) $X = \Bbb Q$, $Y = \Bbb R \setminus \Bbb Q$

(B.) $X = \Bbb Z$, $Y = [-1/2,1/2]$

(C.) $X = (-\infty,100]$, $Y = \{p \in N \mid p \text{ is prime} \}$

(D.) $X=(-\infty, 100]$, $Y = \Bbb Z$

I think option $A$ is correct, since we can get irrationals by summing $X$ and $Y$, but I couldn't find a way to get rationals.

Option $B$ is incorrect since sum will be of the form $.... [-3/2,-1/2], [-1/2,1/2], [1/2,3/2]....$ which covers the whole set $\Bbb R$.

Option $D$ is also incorrect.

Option $C$ seems incorrect by intuition, but I am not sure about $C$. I'm not able to get clear idea about the type of sets option $C$ would form. Any suggestion for $C ?$

  • $\begingroup$ btw use \mathbb Z \mathbbQ \mathbb R otherwise they look like regular sets. $\endgroup$ – qwr Aug 14 at 4:48
  • $\begingroup$ Option C is interesting because most primes are not separated by gaps of 100 but infinitely large gaps do exist $\endgroup$ – qwr Aug 14 at 4:49
  • $\begingroup$ It’s “$ℝ \setminus ℚ$”, not “$ℝ / ℚ$”. The corresponding tex command even is \setminus. $\endgroup$ – k.stm Aug 14 at 5:52

In (A) we cannot write a rational number $x$ as $y+z$ with $x$ rational and $y$ irrational because $z=x-y$ would then be rational. In fact $X+Y$ consist only of irrational numbers.

Your answer for (B) is correct.

(C) and (D) are both incorrect. Given any real number $x$ there is a 'large' prime number $p$ such that $x-y \leq 100$. Hence $x=(x-p)+p \in (-\infty, 100] + P$ where $P$ is the set of all prime numbers.


You are correct about 1). Any irrational + rational will always be an irrational so $\mathbb Q + \mathbb {IR}\subset \mathbb {IR} \subsetneq \mathbb R$.

(I'm using the notation $\mathbb {IR}:= \mathbb R\setminus \mathbb Q$ only because I think it is easy to read in this instance. It might not be a universal symbol.)

And you are correct about 2). The intervals $[n-\frac 12, n+\frac 12]$ "cover" the reals so that for any real $z$ there is some $n\in \mathbb Z$ (possibly two) so that $n-\frac 12\le z \le n+\frac 12$. If we let $x = n$ and $y=z -n$ we get $x \in X$ and $y\in Y$ and $z = X+Y$ so $\mathbb R\subset X+Y$. (And as $\mathbb R$ is our universal set $X+Y = \mathbb R$.

Actually proving that for any real $z$ there is an $n$ so that $x-\frac 1n \le x \le x+\frac 12$ can be proven with the Archimedian principal and isn't a given but... I think that's beyond the scope of the problem.

You are correct about D) but you should give a reason. This is simple. If $z$ is a real number than there is an integer $n: n > z$ then $z-n < 0 < 100$ and so $z-n \in X$ and $n \in Y$ and $z = (z-n) + n \in X+Y$ so $\mathbb R \subset X+Y$.

C) is however correct.

There are an infinite number of primes so for an $z \in \mathbb R$ there is a prime $p$ so that $p > z$. The $z - p < 0 < 100$ so $z-p \in X$ and $p\in Y$ and $z = (z-p) + p = z\in X+Y$ so $\mathbb R \subset X+Y$.

  • $\begingroup$ I think you mean $c$ is incorrect, as question asks, for which $X$ and $Y$, $X+Y ≠ R$. $\endgroup$ – Mathsaddict Aug 14 at 5:56

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