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Indicate whether each of the following statements is true or false and justify your answer.

i) $\forall\ $integers$\ $a$, \exists\ $an integer b such that $a+b=0$.
ii) $\exists\ $an integer $a$ such that $\forall\ $integers $b$, $a+b=0$.

For part (i), I attempted the question using a direct proof:

Let $b=-a$.

Then $a+(-a)=0$. Therefore, this statement is true.

However, I got confused for part (ii). From what I can see, it just seems like the first two parts of part (i) and (ii) is switched in places.

Is there anyone who can help explain this for me? Is there a difference in the meaning of the questions?

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    Think carefully about why the order matters here. Can you imagine a single number $a$ such that no matter which $b$ is given $a+b =0$ always? – 2017-02-25
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    @jameselmore Intuitively, it seems like (i) means that there is one integer b for every integer a that makes the equation valid. For (ii), however, it means that there is one sole integer a for all integers b that makes the equation valid. Is my understanding correct? – 2017-02-25
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    Correct. If you can show that no such $a$ exists then you can claim the second statement is false. – 2017-02-25
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    @jameselmore ah, I get it now! Didn't realize order matters that much. Thank you for your help! – 2017-02-25
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    Just because it is stated doesn't mean it is true. Read what the statement *says*. There exists a number $b$ so that $b + a = 0$. For every $a$. So $b + 1 = 0$ and $b + 15 = 0$ and $b + 2,345,698 = 0$. Is that statement true? Is there such a number? Of course not. So the statement is false. You have to prove it is false. Hint: Let $c \ne d$ then $b + c = 0$ and $b +d = 0$ so $b = -c$ and $b = -d$ so $-c = -d$ so $c = d$. That's a contradiction. – 2017-02-25
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    Another answer to your question: http://math.stackexchange.com/questions/2150079/existential-universal-vs-universal-existential-quantifiers/2150082#2150082 – 2017-02-25
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    @fleablood Oh, I didn't think of using this approach at first. Thank you for your help! – 2017-02-25
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    @EthanBolker Didn't notice that thread, and it's pretty enlightening! Thanks for putting it out here! – 2017-02-25

2 Answers 2

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For part (i) you are correct. In part (ii) the statement is "There exists an $a$ such that for all $b, a+b =0 $

Suppose this statement were true, then let's take $b=1 \Rightarrow a=-1$

If we take $b= 2 \Rightarrow a=-2 $ But $a$ must be the same $\forall b$

This is a contradiction, hence the statement must be false.

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    It's much clearer for me now. Thank you for taking the time to answer my question! – 2017-02-25
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The first statement says that any integer number has an opposite. This is true but you havr to prove it starting from some axiomatic definition of integers, deducing the existence of the element $-a=b$.

The second statement says that there is an integer number that is the opposite of all ather numbers. Tis is false, because if, for the same $a$ we have: $a+b=0$ and $a+c=0$ with $b \ne c$, then we have a contradiction: $$ c=c+0=c+a+b=(a+c)+b=b $$

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    Didn't consider using this approach for the proof. Thank you for your input! – 2017-02-25