1. Solution: By definition, we have\[(-v)+(-(-v))=0\quad\text{and}\quad v+(-v)=0.\]This implies both $v$ and $-(-v)$ are additive inverses of $-v$, by the uniqueness of additive inverse, it follows that $-(-v)=v$.

2. Solution: If $a=0$, we are done.

If $a\ne 0$, then $a$ has inverse $a^{-1}$ such that $a^{-1}a=1$. Hence \[v=1\cdot v=(a^{-1}a)v=a^{-1}(av)=a^{-1}\cdot 0=0.\]Here we use associativity in 1.19 and and 1.30.

3. Solution: Let $x=\dfrac{1}{3}(w-v)$, then \[v+3x=v+3\cdot \dfrac{1}{3}(w-v)=v+(w-v)=w.\]This shows existence. Now we show uniqueness. Suppose, we have another vector $x’$ such that $v+3x’=w$. Then $v+3x’=w$ implies $3x’=w-v$. Similarly, $3x=w-v$. Hence \[3(x-x’)=3x-3x’=(w-v)-(w-v)=0.\]By Problem 2, it follows that $x-x’=0$. This shows uniqueness.

4. Solution: Additive identity: there exists an element $0\in V$ such that $v+0=v$ for all $v\in V$ ; This means $V$ cannot be empty.

5. Solution: If we assume the additive inverse condition, we already showed $0v=0$ for all $v\in V$ in 1.29. Now we assume $0v=0$ for all $v\in V$ and then show additive inverse condition. Since we have $0v=0$ for all $v\in V$, we have \[ v+((-1)v)=1v+((-1)v)=(1+(-1))v=0v=0, \]this means the existence of additive inverse, i.e. the additive inverse condition.

6. Solution: This is not a vector space over $\mathbb R$. Consider the distributive properties in 1.19. If this is a vector space over $\mathbb R$, we will have \[\infty=(2+(-1))\infty=2\infty+(-1)\infty=\infty+(-\infty)=0.\]Hence for any $t\in\mathbb R$, one has\[t=0+t=\infty+t=\infty=0.\]We get a contradiction since zero vector is unique.

## Lost Ghost

3 Feb 2020Q4 is incorrect

## Tianze

20 May 2020Are you saying that the title statement itself is wrong?

## disqus_4MuuHb6gF1

22 Jun 2019The proof of 2 is not complete. See alternative proof below:

Suppose av = 0, with a != 0 and v != 0

Since a != 0, there exists (1/a) in F such that (1/a)a = 1.

So we have:

av = 0

(1/a)av = (1/a)0

((1/a)(a))v = 0 ( by multiplicative associativity and 1.30 respectively)

(1)v = 0

v = 0, contradiction

Hence we conclude that a = 0 or v = 0

## Mohammad Rashidi

22 Jun 2019The proof is complete because if a=0, then it is already true..

## Tianze

20 May 2020The proof of the two of you is equivalent, because there are only four cases to discuss

## stan_comp_neuro

13 Feb 2019I think here gives a better answer for question 2, here you did not address why the a!=0 and V != 0 does not exist stackexchange/2595465

## Mohammad Rashidi

22 Jun 2019It implies that if a!=0, then v=0. If a=0, then it is already true. The proof is complete.

## Paweł Pod

9 Jul 2018An another argument:

8 = 8 + 0 = 8 + (8 + (-8)) = (8 + 8) + (-8) = 8 + (-8) = 8 + ((-8) + (-8)) = (8 + (-8)) + (-8) = 0 + (-8) = (-8), this is the contradiction with the assumption that 8 and (-8) denote distinct objects.

## Maxis Jaisi

25 Apr 2017Alternatively, we could just observe that $ t + infty = infty$ for every $ t in mathbf{R}$ automatically violates uniqueness of additive identity.

## Rong Ou

23 Apr 2017For 6 it also violates associativity:

(infty + infty) + (-infty) = infty + (-infty) = 0

infty + (infty + (-infty)) = infty + 0 = infty

## MASHIAT MUTMAINNAH

18 Jul 2016Hi! Thank you so much for all the solutions, I really appreciate your hard work. Can someone please explain to me the contradiction in the solution Ex 1B question 6?

I just can't seem to understand it. Thank you!

## Mohammad Rashidi

14 Jan 2017Since zero vector should be unique.

## Abu-Yakitori

27 Feb 2017Why 0 + t = infinity + t?

## spindash

28 Feb 2017From the first equation we obtain ∞ = 0 (only using distributive properties).

Whereas in the second equation we get t = 0

That is the contradiction, 0 is not unique.

0+t = ∞+t since by the result of the first equation ∞ = 0. So if 0=∞, 0+t=∞+t.