uncool 194 Posted October 24, 2017 Good. 1*(0 + 0)(as z2) =/= 1*0(as z2) + 1*0(as z2) That means you have broken distributivity. In other words, you have broken the exact way in which addition and multiplication interact. 1 Share this post Link to post Share on other sites

conway 27 Posted October 24, 2017 (edited) Well you will have to show me in an equation how I have broken the distributive property. I do not see it. Besides.....I can manipulate z1 and z2 at will. Any attempt to show the distributive property failing out of a placement of z1 and z2...simply requires that I edit z1 and z2 accordingly. Then.....re apply to the axiom. Only regarding 0 of course. I wait and see.... Edited October 24, 2017 by conway -1 Share this post Link to post Share on other sites

uncool 194 Posted October 24, 2017 Just now, conway said: Well you will have to show me in an equation how I have broken the distributive property. I do not see it. I just did. 1*(0 + 0)(as z2) =/= 1*0(as z2) + 1*0(as z2). 1 Share this post Link to post Share on other sites

conway 27 Posted October 24, 2017 (edited) 3 minutes ago, uncool said: I just did. 1*(0 + 0)(as z2) =/= 1*0(as z2) + 1*0(as z2). Ok....now explain how this fails the distributive property https://www.khanacademy.org/math/pre-algebra/pre-algebra-arith-prop/pre-algebra-ditributive-property/a/distributive-property-explained lol I'm sorry I see your caring it over to both zeros....................you must add both zeros first PEMDOS my friend....lol Edited October 24, 2017 by conway 0 Share this post Link to post Share on other sites

uncool 194 Posted October 24, 2017 I am following PEMDAS (note: the 5th operation is addition, so PEMDAS). To use all the relevant parentheses: 1*(0 + 0)(as z2) =/= (1*0(as z2)) + (1*0(as z2)). It fails distributivity because...that's what distributivity is. I just put in 1, 0, and 0 for the 3 variables. Distributivity in general says that: a*(b + c) = a*b + a*c. That inequality is a counterexample, setting a = 1, b = c = 0. 1 Share this post Link to post Share on other sites

conway 27 Posted October 24, 2017 Im sorry but your not. ANYTHING inside a set of PARENTHESIS must operate first. The addition of ( 0 + 0 ) MUST be performed BEFORE the distributive property. Because it is in the parenthesis. The "distributive" property which is multiplication.... can occur only after what is in the parenthesis is solved. -1 Share this post Link to post Share on other sites

uncool 194 Posted October 24, 2017 (edited) 1 minute ago, conway said: Im sorry but your not. ANYTHING inside a set of PARENTHESIS must operate first. The addition of ( 0 + 0 ) MUST be performed BEFORE the distributive property. Because it is in the parenthesis. The "distributive" property which is multiplication.... can occur only after what is in the parenthesis is solved. Distributivity isn't multiplication. It's the property that the two sides must be equal. And you have agreed that they are not. Edited October 24, 2017 by uncool 1 Share this post Link to post Share on other sites

conway 27 Posted October 24, 2017 (edited) distributive is an act of multiplication.....the parenthesis must be solved before you can multiple in ANY fashion distributive or not. https://www.khanacademy.org/math/pre-algebra/pre-algebra-arith-prop/pre-algebra-ditributive-property/a/distributive-property-explained Edited October 24, 2017 by conway 0 Share this post Link to post Share on other sites

uncool 194 Posted October 24, 2017 1 minute ago, conway said: distributive is an act of multiplication.....the parenthesis must be solved before you can multiple in ANY fashion distributive or not. 1) I did. Remember when I asked you whether 1 * 0(as z2) = 1? And therefore that 1 * (0 + 0)(as z2) = 1? That's because I was following PEMDAS. 2) The distributive property of multiplication is the fact that the two must be equal. No, the distributive property is not an "act" of multiplication. Distribution of multiplication is an "act" that uses the distributive property. To be more precise with that inequality: 1 * (0 + 0)(as z2) = 1 * 0(as z2) = 1 =/= 2 = 1*0(as z2) + 1*0(as z2) The fact that the far left is not equal to the far right means that in your system, distributivity fails. 1 Share this post Link to post Share on other sites

conway 27 Posted October 24, 2017 (edited) Let's assume your right. It does NOT mean it fails. IT means that in the cases involving zero...for the distributive property to hold a specific application of z1 and z2 must occur remember when I said you must acknowledge that 1 * 0 (as z1) = 0 Do you see how I can rewrite the equations following the idea that the distributive property is a matter of equality and not multiplication. Look lets say i do the distributive property first.... 1*(0 + 0)(as z2) = 1*0(as z2) + 1*0(as z2) 1 + 1 = 1 + 1 2 + 2 Perhaps this was a drastic misunderstanding on my part...for that I'm sorry. Edited October 24, 2017 by conway 0 Share this post Link to post Share on other sites

uncool 194 Posted October 24, 2017 OK, that gets into a major problem: It sounds like you're saying your system is "Multiplication gives two answers, and I pick and choose which answer seems to work". That's not how multiplication works, and breaks multiplication entirely. One important thing about multiplication is that it gives a single answer, period. Moreover, one of those answers is simply "Apply the usual multiplication". Which will always work. So your extra answer doesn't add anything. 1 Share this post Link to post Share on other sites

conway 27 Posted October 24, 2017 (edited) 3 minutes ago, uncool said: OK, that gets into a major problem: It sounds like you're saying your system is "Multiplication gives two answers, and I pick and choose which answer seems to work". That's not how multiplication works, and breaks multiplication entirely. One important thing about multiplication is that it gives a single answer, period. Moreover, one of those answers is simply "Apply the usual multiplication". Which will always work. So your extra answer doesn't add anything. Not so... While binary multiplication by zero is RELATIVE...that is it yields two products...each product has a unique solution. Yes further "clarity" is needed as to these unique solutions. I appreciate your help. Edited October 24, 2017 by conway 0 Share this post Link to post Share on other sites

uncool 194 Posted October 24, 2017 Just now, conway said: Not so... While binary multiplication by zero is RELATIVE...that is it yields two products...each product has a unique solution. That's my point - it yields 2 products, and you seem to be willing to pick and choose which one is important for the axioms. Which...isn't useful. Especially because the one that is important for the axioms is always the one that yields the original multiplication. 1 Share this post Link to post Share on other sites

conway 27 Posted October 24, 2017 Just now, uncool said: That's my point - it yields 2 products, and you seem to be willing to pick and choose which one is important for the axioms. Which...isn't useful. Especially because the one that is important for the axioms is always the one that yields the original multiplication. actually I think we had a misunderstanding regarding the axioms....consider if I use the distributive property first as you suggested. 1*(0 + 0)(as z2) =/= 1*0(as z2) + 1*0(as z2) 1 + 1 = 1 + 1 2 = 2 If there is a unique solution to each product where is the problem? 0 Share this post Link to post Share on other sites

uncool 194 Posted October 24, 2017 2 minutes ago, conway said: actually I think we had a misunderstanding regarding the axioms....consider if I use the distributive property first as you suggested. 1*(0 + 0)(as z2) =/= 1*0(as z2) + 1*0(as z2) 1 + 1 = 1 + 1 2 = 2 If there is a unique solution to each product where is the problem? The problem is that you are now not using PEMDAS. The distributive property isn't telling you how to do that multiplication. It's specifying what the result of that multiplication must be. 2 Share this post Link to post Share on other sites

conway 27 Posted October 24, 2017 (edited) 7 minutes ago, uncool said: The problem is that you are now not using PEMDAS. The distributive property isn't telling you how to do that multiplication. It's specifying what the result of that multiplication must be. Look I was under the assumption that you added the two zero's first that is.... 1*(0 + 0)(as z2) =/= 1*0(as z2) + 1*0(as z2) 1 * 0 (as z2) =/= 1 * 0 (as z2) + 1* 0 (as z2) 1 =/= 2 but if I USE the distributive property before adding the two zero's.... 1*(0 + 0)(as z2) =/= 1*0(as z2) + 1*0(as z2) 1 + 1 = 1 + 1 so if I do it as you suggest.....the operations are equivalent....that is I "carry then multiple" the one to both zero's before adding them. Edited October 24, 2017 by conway 0 Share this post Link to post Share on other sites

uncool 194 Posted October 24, 2017 Again: the distributive property is not an act. It is a statement of equality. And the point is that the two are not equal. 1*(0 + 0) means "Do the operation 0 + 0 first, then perform the multiplication 1 * the result". The distributive property says that that must be equal to what you get when you add 1*0 and 1*0. And in your system, it isn't. Of course when you "distribute" 1*(0+0) you get 1*0 + 1*0. That's what distribution says. The point is that in your system, they're not equal, and therefore that the distributive property fails in your system. 2 Share this post Link to post Share on other sites

conway 27 Posted October 24, 2017 13 minutes ago, uncool said: Again: the distributive property is not an act. It is a statement of equality. And the point is that the two are not equal. 1*(0 + 0) means "Do the operation 0 + 0 first, then perform the multiplication 1 * the result". The distributive property says that that must be equal to what you get when you add 1*0 and 1*0. And in your system, it isn't. Of course when you "distribute" 1*(0+0) you get 1*0 + 1*0. That's what distribution says. The point is that in your system, they're not equal, and therefore that the distributive property fails in your system. I was AGREEING with you! OK it is a matter of equality....therefore I must multiple 1 by both 0's in the first set of parenthesis..therefore 1*(0 + 0)(as z2) = 1*0(as z2) + 1*0(as z2) these two expressions are equivalent......because of the distributive property.... -1 Share this post Link to post Share on other sites

uncool 194 Posted October 24, 2017 5 minutes ago, conway said: I was AGREEING with you! OK it is a matter of equality....therefore I must multiple 1 by both 0's in the first set of parenthesis..therefore 1*(0 + 0)(as z2) = 1*0(as z2) + 1*0(as z2) these two expressions are equivalent......because of the distributive property.... Once again: It's not that they simply are equivalent. It's that the distributive property holds only if they are equal. And in your system, they are not equal. 1 Share this post Link to post Share on other sites

conway 27 Posted October 24, 2017 Yes they are equal....if I do NOT add the zero's as I thought I was supposed to. If I do it according to the fashion you suggest then the statements are equivalent. Thanks for your hard work...you deserve the +1's.... 0 Share this post Link to post Share on other sites

uncool 194 Posted October 24, 2017 Just now, conway said: Yes they are equal....if I do NOT add the zero's as I thought I was supposed to. If I do it according to the fashion you suggest then the statements are equivalent. Thanks for your hard work...you deserve the +1's.... You had it right the first time. Once again: the distributive property is a statement of equality, not of notation. 1 Share this post Link to post Share on other sites

conway 27 Posted October 24, 2017 1 minute ago, uncool said: You had it right the first time. Once again: the distributive property is a statement of equality, not of notation. If I had it right the first time then all I have to do is change the application of z2 to z1 on the second half of the equation......for the equation to be equivalent...purely a matter of pinning down a finalized set of axioms.... once again.... 0 Share this post Link to post Share on other sites

uncool 194 Posted October 24, 2017 5 minutes ago, conway said: If I had it right the first time then all I have to do is change the application of z2 to z1 on the second half of the equation......for the equation to be equivalent...purely a matter of pinning down a finalized set of axioms.... once again.... And once again: it seems that you are choosing z1 or z2 in every case to simply match what happens for the original multiplication in order to check the axioms. Which means that your system doesn't add anything useful - in every case that matters, we'd just use basic multiplication. 2 Share this post Link to post Share on other sites

conway 27 Posted October 24, 2017 1 minute ago, uncool said: And once again: it seems that you are choosing z1 or z2 in every case to simply match what happens for the original multiplication in order to check the axioms. Which means that your system doesn't add anything useful - in every case that matters, we'd just use basic multiplication. If the expression... 1*(0 + 0)(as z2) = 1*0(as z2) + 1*0(as z2) must be equivalent... and if I had it right the "first" time and I must add the two zero's.... thereby making my equation NOT equivalent....and all I have to do is state something to the affect of ... "Let any equality possessing multiple zero's consider the following application of z1 and z2" then it is extremely useful.... I will ponder what you have shared with me....thank you. 1 Share this post Link to post Share on other sites

studiot 1359 Posted October 24, 2017 Quote Conway to uncool Thanks for your hard work...you deserve the +1's.... I'll second that +1 to uncool and also +1 to conway for (half) listening. 0 Share this post Link to post Share on other sites