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Suppose f:A->0, where 0 is the empty set and A is non-empty. Using vacuous truth, we can prove f is a bijection.
Surjectivity: "For all y that belongs to 0, there exist a x in A such that f(x)=y"
Since there aren't any y in 0, this proposition is automatically satisfied by vacuous truth
Injectivity: "For all x1 and x1 in A that aren't equal, f(x1) isn't equal to f(x2)"
Since for all x in A, f(x) isn't even defined, again this proposition is satisfied.
I'm not an expert in anything, but this "thing" shows every set has "0 elements". Is there something wrong with my reasoning? Or is it just the strangeness of vacuous truth? —Preceding unsigned comment added by Standard Oil ( talk • contribs) 07:47, 2 July 2009 (UTC)
You are misusing the phrase "vacuous" which if we have a predicate φ(x), where x ranges over X, then ∀x.φ(x) is true when X is empty, even if φ(x) is not true for any x. Instead, if one supposes f:A->0, where 0 is the empty set and A is non-empty, then one can derive 0=1 because the premises are contradictory. This does not establish that 0=1 by vacuous truth, although it does establish that 0=1 for every f that satisfies the premises. — Charles Stewart (talk) 09:13, 2 July 2009 (UTC)
Why does safety stock vary with SQRT(Lead Time) & not Lead Time? —Preceding unsigned comment added by Bharat Kantharia ( talk • contribs) 09:01, 2 July 2009 (UTC)
Hi, I am doing some self study of precalculus.
When factoring polynomials by grouping, you are supposed to change the sign of each term in the if there is a minus sign before the group. This seems wierd to me because I don't see how the act of adding parenthises changes the expression.
To see what I am talking about look at Factorization#Factoring_by_grouping, and see how the signs change in the second group.
I can't find any resources that explain why the signs change, what am I missing here? Thanks for any insight or pointers to more info. -- 99.129.153.2 ( talk) 14:10, 2 July 2009 (UTC)
...and here's an example with a minus sign on the inside becoming a plus sign on the outside:
Say you have $1000 in your checking acount. Something that normally costs $200 is on sale for $200 − $10. After you pay for it with your debit card, your checking account balance is $1000 minus the price, and the price is $200 − $10, so your account balance is now $10 MORE than it would be if you'd paid the regular price. Hence plus rather than minus. Michael Hardy ( talk) 18:02, 2 July 2009 (UTC)
Thanks everyone I think I've got my head wrapped around this now. Great examples! -- 99.129.153.2 ( talk) 19:08, 2 July 2009 (UTC)
Mathematics desk | ||
---|---|---|
< July 1 | << Jun | July | Aug >> | July 3 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
Suppose f:A->0, where 0 is the empty set and A is non-empty. Using vacuous truth, we can prove f is a bijection.
Surjectivity: "For all y that belongs to 0, there exist a x in A such that f(x)=y"
Since there aren't any y in 0, this proposition is automatically satisfied by vacuous truth
Injectivity: "For all x1 and x1 in A that aren't equal, f(x1) isn't equal to f(x2)"
Since for all x in A, f(x) isn't even defined, again this proposition is satisfied.
I'm not an expert in anything, but this "thing" shows every set has "0 elements". Is there something wrong with my reasoning? Or is it just the strangeness of vacuous truth? —Preceding unsigned comment added by Standard Oil ( talk • contribs) 07:47, 2 July 2009 (UTC)
You are misusing the phrase "vacuous" which if we have a predicate φ(x), where x ranges over X, then ∀x.φ(x) is true when X is empty, even if φ(x) is not true for any x. Instead, if one supposes f:A->0, where 0 is the empty set and A is non-empty, then one can derive 0=1 because the premises are contradictory. This does not establish that 0=1 by vacuous truth, although it does establish that 0=1 for every f that satisfies the premises. — Charles Stewart (talk) 09:13, 2 July 2009 (UTC)
Why does safety stock vary with SQRT(Lead Time) & not Lead Time? —Preceding unsigned comment added by Bharat Kantharia ( talk • contribs) 09:01, 2 July 2009 (UTC)
Hi, I am doing some self study of precalculus.
When factoring polynomials by grouping, you are supposed to change the sign of each term in the if there is a minus sign before the group. This seems wierd to me because I don't see how the act of adding parenthises changes the expression.
To see what I am talking about look at Factorization#Factoring_by_grouping, and see how the signs change in the second group.
I can't find any resources that explain why the signs change, what am I missing here? Thanks for any insight or pointers to more info. -- 99.129.153.2 ( talk) 14:10, 2 July 2009 (UTC)
...and here's an example with a minus sign on the inside becoming a plus sign on the outside:
Say you have $1000 in your checking acount. Something that normally costs $200 is on sale for $200 − $10. After you pay for it with your debit card, your checking account balance is $1000 minus the price, and the price is $200 − $10, so your account balance is now $10 MORE than it would be if you'd paid the regular price. Hence plus rather than minus. Michael Hardy ( talk) 18:02, 2 July 2009 (UTC)
Thanks everyone I think I've got my head wrapped around this now. Great examples! -- 99.129.153.2 ( talk) 19:08, 2 July 2009 (UTC)