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I have example in which there is only option of contradiction to prove that identity does not exist.

Consider the binary operation : $a*b= 2a +b, \forall a,b \in \mathbb{Z}$.
Suppose that $\mathbb{Z}$ has an identity $e$ w.r.t. $*$. Then, $e*1 = 2e + 1 = 1 \implies 2e = 0 \implies e=0$. But, $1*0 = 2 + 0 = 2 \ne 1$, so $0$ cannot be an identity.

Sahiba Arora
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jiten
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    A direct proof has to show that for all $a$ there is a $b$ such that either $ab\neq b$ or $ba\neq b$. Take $b(a)=a$ for $a\neq0$ and $b(0)=1$ or many other choices. – orole Jan 12 '18 at 22:29
  • @orole I am not clear with your response's first line example, and hope you meant: "$\forall a$, there $\exists b$ s.t. either $ab \ne a$ or $ba \ne a$". If so, then have you used the concept of left-identity and right-identity. The second line is more clear and takes an identity function as $b()$ with non-zero parameter $a$ s.t. $b(a) \ne a$ for direct proof. Also, I hope by direct proof in-applicability you meant that it is exhaustive and hence for a large or infinite set, not feasible. – jiten Jan 12 '18 at 22:48

2 Answers2

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If you take as an axiom that identities are unique (call it proposition $A$), then there is a unique element $e$ such that $e*x=x$ for all $x\in \mathbb{Z}$ (call it porposition $B_{1}(e)$) and $x*e=x$ for all $x\in \mathbb{Z}$ (call it proposition $B_{2}(e)$).

In symbols: $A\Rightarrow (\exists e\in \mathbb{Z}(B_{1}(e) \wedge B_{2}(e)))$

Equivalently: $(\forall e\in \mathbb{Z}(¬B_{1}(e) \vee ¬ B_{2}(e)))\Rightarrow ¬A$

In other words, if you prove that for every element $e\in \mathbb{Z}$ such that $e*x=x$ for all $x\in \mathbb{Z}$ there exists an element $y\in \mathbb{Z}$ such that $y*e \neq y$, then you can deduce that an identity does not exist.

Warning: this is not a proof by contradiction, it is a proof by contrapositive. We used the implication $A\Rightarrow B$ in its equivalent form $¬ B \Rightarrow ¬ A$. But if I remember the logic course I took correctly, to prove this equivalence you are actually using the falsum deduction rule. So you are doing some sort of proof by contradiction implicitly.

If you are interested you can read this question.

Pedro
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$e \in \mathbb Z$ is identity if for every $a \in \mathbb Z,$ $$a*e=e*a=a.$$

To prove directly that $\mathbb Z$ doesn't have identity we need to show that no element of $\mathbb Z$ is identity. So we need to show that no element of $\mathbb Z$ satisfies the above equalities for every $a.$

That means we need to show that for every $x \in \mathbb Z,$ there exists $a \in \mathbb Z$ such that either $x *a \neq a$ or $a *x \neq a.$

Let $x \in \mathbb Z.$

If $x =0,$ then take $a=1.$ So $a*x=2 \neq a.$

If $x \neq 0,$ then take $a=x.$ So, $x*a=2x+a \neq a.$

Sahiba Arora
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  • I hope that am confused in the main para., you seem to consider here, many identities $x$. So, can there be many identities. Also, there is some confusion regarding the values possibly taken by $a$, can it take value of identity, i.e. $x$ or $e$. – jiten Jan 12 '18 at 23:37
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    I'm not considering many identities. I'm just taking an arbitrary element of $\mathbb Z$ and showing it is not an identity. I don't understand your last statement. – Sahiba Arora Jan 12 '18 at 23:38
  • Thanks for your edited answer to make it more clear. Just a small confusion (that might be not logical even) - So, when you consider that for every element $x$ (arbitrary, as you stated just now) need prove that $xa = ax \ne a$, then are doing an exhaustive search for a failure case. Is this not different from having an identity before-hand and proving its failure. Also, by this way of exhaustive search, will consider $x = a$ as well, as $a$ can be identity too. – jiten Jan 12 '18 at 23:44
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    There are a lot of times when a direct proof and a proof by contradiction are just re-wordings of each other. This happens a lot when you have to show something does not exist. You either assume it exists and arrive at a contradiction or show that no element is that something. Mostly you'll just be rephrasing the argument. – Sahiba Arora Jan 12 '18 at 23:47
  • Very interesting to know/realize that. My ignorance also stems from the fact that usually (in literature, or elsewhere) there is proof by only contradiction based approach, or contra-positive one. – jiten Jan 12 '18 at 23:52
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    That's because of the reason I stated above. Since direct proof will only be a re-wording, mostly the proof by contradiction would require fewer words and can be said more elegantly. – Sahiba Arora Jan 12 '18 at 23:56