Recent content by RM86Z

Ah brilliant that makes sense now thank you so much PeroK! That was a very slight detail that I wasn't seeing. I appreciate the help.

When ## x_1 - x_2 = 0 ## and actually now that I think about it it wasn't defined in the hypothesis that ## b,d \neq 0 ## as that certainly isn't valid either? EDIT: I suppose the last bit about ## a,b \neq 0 ## doesn't make sense as it depends on the ordering of the algebra. In other words...

Sorry I know there is a connection here I'm just not getting it at this stage. By the properties of the determinant the result ## ad - bc = 0 ## tells me that the two rows of the linear system are the same so ## a = b ## and ## c = d. ##

I don't think it is a proof as I need to prove that ## (x_1,y_1) ## and ## (x_2,y_2) ## are equal, or in other words ## x_1 - x_2 = 0 ## and ## y_1 - y_2 = 0. ## I cannot see how to get there from the result I have ended with? I have done some linear algebra a long time ago but I am very rusty...

Given ## a,b,c,d,e,f \in \mathbb {R}, ad - bc \neq 0 ##, if ##(x_1,y_1)## and ##(x_2,y_2)## are pairs of real numbers satisfying: ## ax_1 + by_1 = e, cx_1 + dy_1 =f ## ## ax_2 + by_2 = e, cx_2 + dy_2 = f ## then ## (x_1,y_1) = (x_2,y_2). ## Here is my attempt at a proof, I have gotten stuck...

Nice one thank you PeroK

Awesome thank you

Ah yes of course thanks for pointing that out - is this better? 1) given rational numbers s,t with t != 0 2) s = a/b, t = c/d for integers a,b,c,d with b,c,d != 0 3) (a/b) / (c/d) = (a/b) * (d/c) = ad / bc is a rational number as ad and bc are integers (product of integers is an integer) 4) s =...

Is my proof correct? The steps from hypothesis to conclusion are in order below: 1) given rational numbers s,t with t != 0 2) take s = p/1 and t = q/1 where p,q are integers 3) (p/1)/(q/1) = p/q is rational 4) therefore by substitution s/t is rational

Ah brilliant I understand now: 32 cases for each of {1,2},{1,3} and {2,3} for a total of 96. 243 - 96 = 147. Then add three of the cases 11111, 22222, 33333 back in. And vela thank you I see that now! It is definitely much simpler to exclude those cases, I was thinking the other way around...

Thank you guys, I think perhaps I am struggling a little with how to count the copies. Delta2, you say that there are 30 cases of 4 copies of one digit. I calculate 45? Taking four copies of 1: 1111x 111x1 11x11 1x111 x1111 There are three choices for x (1,2 or 3), so this is 5 x 3 = 15. Then...

Thanks guys I think I've got it now? case 1: only 1’s eg. 1,2,3,1,1 -> (60 x 1)/3! = 10 * case 2: only 2’s eg. 1,2,3,2,2 -> (60 x 1) /3! = 10 * case 3: only 3’s eg. 1,2,3,3,3 -> (60 x 1)/3! = 10 * * only one way to choose from those case 4: 1,2 eg. 1,2,3,1,2 -> (60 x 4)/(2! x 2!) = 60 # case...

Thanks PeroK I am still struggling with this I have broken it down into a smaller case of a set with 3-digit integers, each digit is either of 1 or 2: The total number of digits is 2^3 = 8. And the number of digits containing both 1 and 2 is 6 by 3P2 = 3!/1! = 6. The list is as follows: 1,1,1...

Perhaps it should be: (5P3 * 3^2) / (3! * 2! * 1!) = 90 (by the multinomial theorem)?

Thank you I didn't think of it that way but that does make sense!