-1

Background:

Exercise 5: If $a=bc$ with $a\neq 0$ and $b$ and $c$ nonunits, show that $a$ is not an associate of $b$

Questions:

For the above question, is the contrapositive of the statement of the above exercise: if $a$ is an associate of $b$, and $a=bc$ with $a\neq 0$ then $b$ or $c$ is a unit.

I am haveing trouble with the latter part of the statement which has the following:

$a=bc$

$a\neq 0$

$b$ is a not a unit

$c$ is a not a unit

$b$ is a not a unit, $c$ is a not a unit, I can negate it to become a disjunction. But "$a=bc$ with $a\neq 0$", do I treat the "with" here as an and, or do i simply consider "$a=bc$ with $a\neq 0$" as one single statement?

Thank you in advance

KReiser
  • 65,137
Seth
  • 3,325
  • Some major mathjax is missing -- your proposed converse is unreadable – coffeemath Feb 05 '24 at 03:17
  • @coffeemath I fixed it. I was missing a dollar sign. – Seth Feb 05 '24 at 03:18
  • I think you need to replace the tail end "and either b or c are units." by "then either b or c is a unit. – coffeemath Feb 05 '24 at 03:39
  • @coffeemath do you mean like this: "For the above question, is the contrapositive of the statement of the above exercise that $a$ is an associate of $b$ then if $a=bc$ with $a\neq 0$ then either $b$ or $c$ is a unit." – Seth Feb 05 '24 at 03:44
  • 1
    Be careful how you use the word "if." You started with a statement in the form "if P then Q" and you seem to be trying to write "not Q then if not P." That's not a contrapositive; it's not even a grammatical sentence. You should be writing "if not Q then not P." – David K Feb 06 '24 at 08:13
  • @DavidK I edited the statement of the contrapositive, I think the portion "$a=bc$ with $a\neq 0$, I am not sure if that is treated as a separate statement or do I treat it as part of $b$ and $c$ are nonunits. – Seth Feb 06 '24 at 16:38
  • I think you have a true statement now that's derivable from the contrapositive, but not the same thing. You seem to be making this exercise more complicated than it ought to be. Identify P and Q in the original statement so that the statement is "if P then Q"; then write "if not Q then not P". The only additional rewriting of the sentence should be to phrase "not Q" and "not P" in a more natural way, but not to shuffle little bits of P or Q from the "if" part to the "then" part or vice versa. – David K Feb 06 '24 at 17:41
  • @DavidK I edit my post to reflect why I was having trouble with writing out the contrapositive. There is a grammartical issue of dealing with the word "with". I am not sure how to symbolize that in therms of symboblic logic. – Seth Feb 06 '24 at 21:10
  • That clarifies the question immensely, thanks. I agree with the answer that was posted yesterday; in this case, "with" is a stylish way of saying "and". – David K Feb 06 '24 at 23:49
  • 2
    Stop using MathJax to format text, please. Use markdown. This is very far from the first time you've been told this. – KReiser Feb 08 '24 at 05:24

1 Answers1

2

Given:

((= with ≠0) and ( and nonunits)) $\rightarrow$ ( is not an associate of ).

The contrapositive is:

$\neg$ ( is not an associate of ) $\rightarrow$ $\neg$ ((= with ≠0) and ( and nonunits)).

$\neg$ ( is not an associate of ) $\rightarrow \neg$ ((= $\land$ ≠0) $\land$ ( nonunit $\land$ nonunit))

( is an associate of ) $\rightarrow$ ($\neg$(= $\land$ ≠0) $\lor$ $\neg$( nonunit $\land$ nonunit))

( is an associate of ) $\rightarrow$ $\neg$(=) $\lor$ $\neg$(≠0) $\lor$ ( unit $\lor$ unit)

( is an associate of ) $\rightarrow$ ($\neq$) $\lor$ (=0) $\lor$ ( unit $\lor$ unit)

( is an associate of ) $\rightarrow$ ($\neq$) $\lor$ (=0) $\lor$ ( unit) $\lor$ ( unit)

I would write: If is an associate of then $a\neq bc$ or $a=0$ or $b$ is a unit or $c$ is a unit.

I don't know the actual content of your statement so it may make sense to stop applying the negations earlier to have your contrapositive statement that you would like to prove. The statement as you wrote it is missing some if/then structure.

RobinSparrow
  • 1,091
  • if $a$ is an associate of $b$ then $a=bc$ with $b$ is a unit or $c$ is a unit? I mean it doesn't contradict the part where you state "$a\neq bc$ or $a=0$ or $b$ is a unit or $c$ is a unit." – Seth Feb 05 '24 at 04:09
  • I'm confused by you saying with. The statement evaluates as true if, provided that $a$ is an associate of $b$, that you can prove that one of the 4 statements separated by ORs after the arrow hold. I'll edit to make the 4 possibilities clearer since I kept the last two together. – RobinSparrow Feb 05 '24 at 04:13
  • usually with "$b$ is an associate of $a$ means that $a=bc$ for some unit $c.$ I am just wondering if $b$ or $c$ is a unit, that would mean that $a=bc$ – Seth Feb 05 '24 at 04:15
  • If, when considering the contrapositive, you can prove that c is a unit, then you satisfy the OR after the arrow, and the statement is true. Since it is true, its contrapositive, which is your original statement, is also true. You only discover things about $a=bc$ in the contrapositive direction, and in the original direction, you assume it is so. – RobinSparrow Feb 05 '24 at 04:19
  • ah so in the contrapositive if I assume that $a$ is an associate of $b$ and show that $c$ is a unit, then I have proved the contrapositive statement. And hence taking the contrapositivev again, gives $a-bc$. That is what you maen? – Seth Feb 05 '24 at 04:22
  • First sentence: yes you satisfy the statement, making both the contrapositive and the original true, with all their implications. Since I don't know the topic you're studying, I'm not able to say much about what you can traditionally assume or would be driving at with your proof. I was trying to illustrate the creation of the contrapositive and passing the negation through various AND and OR statements so you see how it's done. I immediately edited my prior comment when I realized I had misread the sentence, and your second sentence is probably off: in the original statement, we assume $a=bc$. – RobinSparrow Feb 05 '24 at 04:30
  • Okay thank you. – Seth Feb 05 '24 at 04:32
  • I edited my question. I was wondering if i can ask you something quick. When i see a statement in propositional calculus or in a math statement, how do I intepret the word "with". Do i consider that as an "and"? – Seth Feb 06 '24 at 21:45
  • 1
    I view it as an “and”. – RobinSparrow Feb 06 '24 at 21:57
  • ah kk. thank you. It is these little grammar points that trips me up. – Seth Feb 06 '24 at 22:01