Let $F$ be a field. (a) If $1 + 1 = 0$, show that $a + a = 0$ for all $a \in F$. (b) If $a + a = 0$ for some $a \neq 0$, show that $1 + 1 = 0$

Question

Let $F$ be a field.

(a) If $1 + 1 = 0$, show that $a + a = 0$ for all $a \in F$.

(b) If $a + a = 0$ for some $a \neq 0$, show that $1 + 1 = 0$

I have found proof's for $1+1=0$ but I am not sure if it is the right proof for this question. I am a bit unsure as to that the question is asking, do I assume $F$ is a field $\{0,1\}$? anyone who can show me the answer and how to do it would be greatly appreciated.

Answer 1

Hint for (a): Given $1+1=0,$ multiply by $a$ and apply distributive law. Also need that $a\cdot 0=0$ for any $a$ but that should be a lemma early on [also can be proved fairly easily from field axioms].

Answer 2

It is easy provided by the field axioms; specifically, if $a,b,c\in F$ where $(F,+,\cdot)$ is a field, then $a\cdot(b+c)= a\cdot b+a\cdot c$. Using this we have

$$1+1=0$$ $$a\cdot(1+1)=a\cdot 0$$ $$a\cdot 1+a\cdot 1=0$$

$$a+a=0$$

Note that all these steps are reversible, because in a field all non-zero elements are units, i.e., have an inverse, and for it to be a field it must be a domain so cancellation law must hold. Since all steps are reversible, this also answers $(b).$

Answer 3

It's very easy to show the given relations . And we are gonna work with some basic properties of a field. Let( F ,+, .) Is our field . Now it's given, 1+1=0. Now if , a belongs to F then by using the property, a.(b+c)=a.b+a.c , for all , a,b,c belongs to F , we can write,

a.(1+1)=a.1+a.1 = a+a , as '1' is multiplicative identity. So, a+a=0 ,as a.0=0 ....(p)

[ a.0= 0 because, 0+0=0 , as 0 is additive identity. Now if a belongs to F then, a.0 +a.0=a.0 Now as a belongs to F , -a belongs to F , as it's the additive inverse. So , -a.0+a.0+a.0=-a.0+a.0 Or, a.0=0 ]

Now for some a≠0 , if a belongs to F, the (1/a) belongs to F as (1/a) is multiplicative inverse of a . Now , as a+a=0 , by using a.(b+c)=a.b+a.c , we can write (1/a).(a+a)= a.1/a+a.1/a . As, a.1/a=1 , then

1+1=0.

Answer 4

I am a bit unsure as to that the question is asking, do I assume $F$ is a field $\{0,1\}$?

Let's read through the requirements of the problem carefully. All that is given to us is: "Let $F$ be a field." So we know $F$ is a field -- we don't know how many elements it has, but all fields at least have a zero element (which is written $0$) and a one element (which is written $1$).

(a) If $1 + 1 = 0$, show that $a + a = 0$ for all $a \in F$.

Now for this part, it does not assume that $F = \{0,1\}$ like you said; but it's just stating a fact about $0$ and $1$, which are elements of the field. So it is stating the fact that $1 + 1 = 0$. Then we have to prove that $a + a = 0$, for any $a \in F$. To do this, try multiplying by $a$ and using the field axioms.

(b) If $a + a = 0$ for some $a \neq 0$, show that $1 + 1 = 0$

For this part, try "dividing" by $a$ instead of multiplying. What axiom of the fields would let us divide by $a$? Remember that "dividing" really means multiplying by the multiplicative inverse.