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Mathematics 12 Online
OpenStudy (akashdeepdeb):

Can someone help in this question?

OpenStudy (akashdeepdeb):

OpenStudy (hoblos):

if we choose the color (bc) a*bc = bc rule iv b*bc = bc rule i c*bc = bc *c rule ii = bc

OpenStudy (akashdeepdeb):

Can you explain the question once? I have no idea about the question.

OpenStudy (hoblos):

do you understand the rules ?

OpenStudy (akashdeepdeb):

Sort of, but I guess first I got to understand the question. :/

OpenStudy (hoblos):

you have to find a color "x" such that ax=x ; bx=x & cx=x

OpenStudy (akashdeepdeb):

What does it mean when they say, x and y are among a, b & c?

OpenStudy (akashdeepdeb):

Does among mean, like a mixture of the colors? |dw:1394481101854:dw|

OpenStudy (hoblos):

well it also can be one of the colors not only a mixture.. it means x and y are from these colors, one color or a mix of them

OpenStudy (akashdeepdeb):

So x and y can be any of the subsets of a,b,c ?

OpenStudy (akashdeepdeb):

And even xy is a subset of a,b,c?

OpenStudy (hoblos):

it is better to say a combination instead of subset, but if you mean that yes xy is also a combination of a,b,c

OpenStudy (akashdeepdeb):

Okay, got it. Now we have to find a combination of a,b,c that follows the rules and gives the same color yeah?

OpenStudy (hoblos):

any combination you find follows the rules.. you must use these rules to prove (ax=x ; bx=x , cx=x)

OpenStudy (akashdeepdeb):

So there you took. x = bc yeah?

OpenStudy (hoblos):

yes

OpenStudy (akashdeepdeb):

a*bc = bc [by fourth rule] b*bc = bc [HOW?] c*bc = bc [HOW?] O.O

OpenStudy (hoblos):

b*bc = bbc bb=b bbc=bc

OpenStudy (hoblos):

c*bc = bc*c = bcc = bc

OpenStudy (akashdeepdeb):

"b*bc = bbc bb=b bbc=bc" xx = x [from first rule] but here, x = bc! [how is bb = b?]

OpenStudy (hoblos):

they used x to generalize the rules, but you can use them for any color you are using

OpenStudy (akashdeepdeb):

But here when we are considering x to be bc Do we not have to go with x = bc everywhere? if x = bc then bc*bc = bc [from first rule] but b*b \(\neq\) b [because x = bc already] I am not really getting it.

OpenStudy (hoblos):

but if we read what they say between brackets "mixing a color with itself results in that color" they said (a color) which means any color not only x

OpenStudy (akashdeepdeb):

Oh okay! And do x and y and z also have to be any combination of a,b and c?

OpenStudy (hoblos):

yes

OpenStudy (akashdeepdeb):

Can you stay online for like 15 more minutes or more? I NEED to submit this by today, I'll just try to do it once on my own now. Thanks!

OpenStudy (hoblos):

okay i'm here if you need any help :)

OpenStudy (akashdeepdeb):

By the way, how did you select 'bc' directly? Any method? Or a random guess?

OpenStudy (hoblos):

i found it logical and tried to prove it, it can be considered as random guess

OpenStudy (akashdeepdeb):

Your method is 100% right. But I just want to know, if I can also guess like that for my submission.

OpenStudy (hoblos):

well we can say to satisfy ax=x x can be any color to satisfy bx=x x must contain b to satisfy cx=x x must contain c so x must be a combination of b and c

OpenStudy (akashdeepdeb):

Why must 'x' contain b in bx? O.O

OpenStudy (hoblos):

so that the b is eliminated we must have bb=b so only the b included in x remains

OpenStudy (akashdeepdeb):

That makes a lot of sense, thanks a lot man, I'll just do this on fresh sheet of paper in some time. :D

OpenStudy (hoblos):

glad to help :D

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