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A relation on the real numbers is given by $a^2 +b^2 =0$. I know this is not reflexive, I believe it is symmetric(although there is only one solution) but I'm not sure whether it can be called transitive as the only solution is $a=b(=0)$.

miracle173
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Anshuman Agrawal
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    If the relation is $a$ ~ $b$ if $a^2+b^2=0$, then as you noted if $a$ ~ $b$ and $b$ ~ $c$ then $ a = b = 0$ and $ b = c = 0$ so $a^2 + c^2 = 0^2 + 0^2 = 0$ thus $a$ ~ $c$, so it is transitive. – Tanny Sieben Jun 14 '21 at 11:26
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    What is $a$ and $b$ here @anushuman . –  Jun 14 '21 at 11:26
  • Thanks anshuman. You can consider @tanny comment. As long as real numbers are concerned only 0R0 is possibility. So others two relation except reflective is true here –  Jun 14 '21 at 11:42
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    Per the comment of @AmitMittal, the transitivity breaks down when $a,b,c$ are permitted to be complex numbers. Example: $a = i, b = 1, c = i.$ Then $a^2 + b^2 = (-1) + (1) = 0$. Similarly, $b^2 + c^2 = 0.$ However, $a^2 + c^2 = -2.$ – user2661923 Jun 14 '21 at 12:37
  • In the question, a and b are to be real numbers (.. Is a relation on the set of real numbers) – Anshuman Agrawal Jun 15 '21 at 12:24

1 Answers1

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The definition of the "the relation is transitive" is

For all $a, b, c$ it holds that if $a\sim b$ and $b\sim c$, then $a\sim c$.

The "if" and "then" here are supposed to be understood in the particular mathematical jargon (known as a "material conditional") where their meaning is that whenever you have a situation where [$a\sim b$ and $b\sim c$] is false, that alone will cause the whole if-then claim to be considered true.

That is completely the case for your relation. Some examples:

$$ \begin{array}{|c|c|}\hline a & b & c & a \sim b\text{ and }b\sim c? & a \sim c? & \text{so, are we happy?} \\ \hline 0 & 0 & 0 & \text{yes} & \text{yes} & \text{yes} \\ \hline 0 & 0 & 1 & \text{no: }b\sim c\text{ is false} & \text{no} & \text{yes} \\ \hline 2 & 2 & 2 & \text{no, neither} & \text{no} & \text{yes} \\ \hline 0 & 5 & 0 & \text{no, neither} & \text{yes} & \text{yes} \\ \hline \end{array} $$

The one combination that can't possibly arise is $$ \begin{array}{|c|c|}\hline a & b & c & a \sim b\text{ and }b\sim c? & a \sim c? & \text{so, are we happy?} \\ \hline ?? & ?? & ?? & \text{yes} & \text{no} & \textbf{no} \\ \hline \end{array} $$ and since that is impossible, your relation is transitive.

For what's worth, there's a similar variety of intermediate results when we check whether $<$ is transitive -- which it had better be, since it's one of the ur-examples of transitive relations:

$$ \begin{array}{|c|c|}\hline a & b & c & a < b\text{ and }b < c? & a < c? & \text{so, are we happy?} \\ \hline 1 & 2 & 3 & \text{yes} & \text{yes} & \text{yes} \\ \hline 3 & 2 & 1 & \text{no, neither} & \text{no} & \text{yes} \\ \hline 1 & 1 & 2 & \text{no: }a<b\text{ is false} & \text{yes} & \text{yes} \\ \hline \end{array} $$

But there's no requirement that all three situations in fact arise. For example the relation that never holds is transitive, since the only situation we can get is

$$ \begin{array}{|c|c|}\hline a & b & c & a \sim b\text{ and }b\sim c? & a \sim c? & \text{so, are we happy?} \\ \hline \text{whatever} & \text{whatever} & \text{whatever} & \text{no, neither} & \text{no} & \text{yes} \\ \hline \end{array} $$

And the relation that relates everything to everything is also transitive since for it we end up with $$ \begin{array}{|c|c|}\hline a & b & c & a \sim b\text{ and }b\sim c? & a \sim c? & \text{so, are we happy?} \\ \hline \text{whatever} & \text{whatever} & \text{whatever} & \text{yes} & \text{yes} & \text{yes} \\ \hline \end{array} $$

Troposphere
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  • Thanks! I've only recently learnt about types of relations so I was somewhat confused but going through your comment a few times I got it. Although I have one more type of question I'm unsure about ( I think you answered a similar question in your comment but I'll ask anyway as I'm not sure they're the same) if a relation is such that if a is related to b, then b can't be related to any other element, ex:- R = {(x, y) : y=x+5 and x<4 } is a relation on N. I initially thought it is not transitive but in the answers it is given to be transitive, but online ncert solutions said it is not – Anshuman Agrawal Jun 15 '21 at 12:58
  • @AnshumanAgrawal: "a relation is such that if a is related to b, then b can't be related to any other element" will necessarily be transitive because this condition says that "$a\sim b$ and $b\sim c$" is always false when $b\ne c$, so we'll always be happy in that case. The desription leaves open the possibility that $b\sim b$ might happen, but if/when that is the case, "$a\sim b$ and $b\sim b$" obviously still implies $a\sim b$, so we're still good. – Troposphere Jun 15 '21 at 13:05
  • @AnshumanAgrawal i am being fan of your struggle and self learning habit. Let me clear to you. ( X , Y ) is set of ordered pair where X is a natural number less than 4. So x can be 1 , 2 , 3. And $Y$ is always greater than x by 5. So possible values of $Y$ corresponding to $X$ is 6 , 7 , 8 . So what are the possible order pair comes , let's count : 1). If x is 1 , y is 6 : so , (1 , 6 ) is first ordered pair . 2). If x is 2 , y is 7 : ( 2 ,7 ) . 3). Similarly ( 3 , 8 ) is last one. So it's not reflective , symmetric . But it's transitive –  Jun 15 '21 at 22:13
  • because 2 related 7 but 7 isn't related to anything then and it's happening with every element so , as per @Troposphere explained if there was 2R7 and 7 was also related to some other no. X then 2 must be related to X for fulfilling the transitivity case. But here b is not related to anything for all b so transitivity is naturally here –  Jun 15 '21 at 22:16
  • Ok, so I learned a bit more, is this question related to Vacuous Truth? – Anshuman Agrawal Oct 25 '21 at 14:56