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I'm just a little confused, I thought the identity for multiplication is always 1, yet I was looking at this problem online and it says that 0 is the identity for this problem, this was taken from "Math is fun":

"Show that the set {0} with multiplication is a group. For any elements a and b of {0}, (a*b) is an element of {0}. The closure law has been followed."

"For any a,b,c of {0}; a*(bc) = (ab)c. The associative law has been followed. For any a of {0} ia=a, where i is a particular element in {0}.The left identity element i is 0 here.

For any a of {0} the equation x*a=i has a solution known as the left inverse of a.0 is the only element in {0} and the left inverse of 0 is 0. All these properties are followed by this set that is closed under multiplication.

Therefore, {0} is a group with respect to multiplication."

This was someones answer to the problem, and I found it very confusing, and I did read up on the theory behind it.

Please advise

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  • $\begingroup$ The only element you have is $0$. Does $0$ respects all of these properties? $\endgroup$ – Alberto Andrenucci Jun 19 '17 at 10:48
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    $\begingroup$ So? I don't understand where the problem is. Explain better so I can help. $\endgroup$ – Alberto Andrenucci Jun 19 '17 at 10:50
  • $\begingroup$ Sorry, basically, I thought the identity is always 1 for multiplication, but it says above in the answer I found that the identity is 0....why? $\endgroup$ – Michael O' Driscoll Jun 19 '17 at 10:52
  • $\begingroup$ @MichaelO'Driscoll Because in this situation the only thing you can multiply with is $0$ and then always $0x=x$ (since $x=0$ always). $\endgroup$ – skyking Jun 19 '17 at 10:57
  • $\begingroup$ Since the problem seem to be that you're confused you should probably explain what confuses you. Otherwise it would be hard for us to unconfuse you. $\endgroup$ – skyking Jun 19 '17 at 10:59
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Closure:

Is it true that in the set $\{0\}$, for any two elements $a,b\in\{0\}$, the product $a\times b$ is also an element of $\{0\}$?

Answer:

Yes! Proof:

  • If $a\in\{0\}$, then $a=0$
  • If $b\in\{0\}$, then $b=0$.
  • Therefore, $a\cdot b=0\cdot 0=0$.
  • Therefore, because $0\in\{0\}$, we conclude $a\cdot b\in\{0\}$.

For associativity, a very similar argument can be made.


Identity:

Is $0$ the identity of $\{0\}$? That is, is it true that for any element $a\in\{0\}$, the element $a\cdot 0=a$?

Answer:

Yes! Proof:

  • If $a\in\{0\}$, then $a=0$.
  • Therefore, $a\cdot 0 = 0\cdot 0=0$.
  • Since $a=0$ and $a\cdot 0=0$, we conclude $a\cdot 0=a$.
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  • $\begingroup$ Thanks, that makes sense, but what about the identity? I would assume it is 1 immediately $\endgroup$ – Michael O' Driscoll Jun 19 '17 at 10:55
  • $\begingroup$ @MichaelO'Driscoll See my added answer. $1$ cannot be the identity because $1$ is not an element of $\{0\}$. $\endgroup$ – 5xum Jun 19 '17 at 10:56
  • $\begingroup$ I see, apologies...just learning this stuff by myself, thanks for your answer $\endgroup$ – Michael O' Driscoll Jun 19 '17 at 10:57
  • $\begingroup$ @MichaelO'Driscoll Don't apologize, you asked a legitimate question. It's designed to be a little confusing, but it teaches an important point (i.e., that sometimes, $0$ can be a multiplicative identity, as long as $0$ is the only number you can multiply by). $\endgroup$ – 5xum Jun 19 '17 at 10:59
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1 and 0 are just names that we use to communicate. It doesn't really matter how you call your elements, it only matters what they do.

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