Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Could I get some feedback on my work below? Thanks in advance.

  • $G = \langle \mathbb{R}, + \rangle , H = \{ x \in \mathbb{R}: \tan x \in \mathbb{Q} \}$

(i). If $\tan a$ and $\tan b$ $\in H$, then from trigonometry \begin{align*} \tan a + \tan b = \tan(a + b) (1 - \tan a \tan b) \notin H \end{align*} For example, if we let a = 0$^{\circ}$ and b = 90$^{\circ}$, then $\tan a + \tan b = \infty \notin \mathbb{Q}$.

Thus $H$ must NOT be a subgroup of $G$.

  • Let $C$ and $D$ be sets, with $C \subseteq D$. Prove that $P_{C}$ is a subgroup of $P_{D}$ where $P_{C}$ and $P_{D}$ are the power sets of C and D respectively

$C \subseteq D$ means that every element in $C$ is an element in $D$. And the operation in this case is the symmetric difference $(\Delta)$ where the nullset is the identity and each element is its own inverse.

In order to show that $P_{C}$ is a subgroup of $P_{D}$:

(i). Let $A$ and $B$ be any two sets in $P_{C}$. The symmetric difference $A \Delta B$ must also be a set in $P_{C}$. (ii). Let $A$ be a set in $P_{C}$. Then the inverse of $A$ must also be in $P_{C}$.

(i). The symmetric difference of two sets $A, B \subseteq P_{C}$ is defined as $A \Delta B = (A \setminus B) \cup (B \setminus A)$. Since $(A \setminus B)$ and $(B \setminus A)$ will always yield sets that are contained in $P_{C}$, we can conclude that $P_{C}$ is closed under the operation of symmetric difference.

(ii). Since the inverse of any set $A$ is itself, then the inverse of every element must also be contained in $P_{C}$. So $P_{C}$ is closed under inverses.

Thus $P_{C}$ is a subgroup of $P_{D}$

share|cite|improve this question
Doublecheck your first counterexample. Why is it important that $a$ and $b$ are in $H$? And are they? – Adam Saltz Aug 9 '11 at 2:58
Your specific example of $a=0^\circ,b=90^\circ$ is invalid because $90^\circ\not\in H$. Use instead $45^\circ\in H, 45^\circ+45^\circ\not\in H$ to show $H$ isn't closed under addition. – anon Aug 9 '11 at 3:05
@IAmBrianDawkins: Yes I see my mistake now. My original examples were not in $H$ the first place. – Student Aug 9 '11 at 3:14
What is $P_C$? Is that notation for the power set of $C$? – Gerry Myerson Aug 9 '11 at 3:18
The second proof looks good to me, with the small omission that you have to note that $P_C$ is non-empty. – Gerry Myerson Aug 9 '11 at 3:35
up vote 3 down vote accepted

For the first question, I am not sure why you computed $\tan a + \tan b$ ? If you want to check that $H$ is closed under addition (which is required to be a subgroup), then you have to check whether the sum two elements still belongs to the set $H$.

If you have two elements $x,y\in H$, then $\tan x \in \mathbb{Q}$ and $\tan y \in \mathbb{Q}$. Now you want to check if $x+y \in H$, which means that you want to check whether $\tan (x+y) \in \mathbb{Q}$. You should be able to use the same identities as you already did to solve this though.

share|cite|improve this answer
If I have two elements tan x and tan y in H, don't I want to see if tan x + tan y is in H and not tan(x + y)? – Student Aug 9 '11 at 4:18
You are absolutely right. I originally thought of $x$ representing $\tan x$ instead of viewing it as a real number. – Student Aug 9 '11 at 4:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.