2 shapes are the same colour
2 shapes are filled
2 shapes have four sides
the point of this is that there are multiple distinct properties not found in any member of the remainder set
sp3ctr4l@lemmy.dbzer0.com 19 hours ago
… only one choice is green.
How is this difficult, other than if you are r/g colorblind?
The correct choice is C.
If you pick A, B is also red, and C is also an irregular 4-gon. So A is not unlike either B or C.
If you pick B, A is also red, and C is also filled solid with color. So B is not unlike either B or C.
But if you pick C, while C does have elements in common with A and B, it is also unlike each of them singly, as well as both of them together, in that it is green.
C is the only choice where ‘is unlike both of them’, them being the other choices… is true, in any sense.
It has a distinct property.
canofcam@lemmy.world 18 hours ago
sp3ctr4l@lemmy.dbzer0.com 18 hours ago
No.
You are wrong.
“Select the image that is unlike the other two.”
The only possible choice that results in a set of 2, and a set of 1, which are seperated cleanly by a distinct property, is picking C.
The goal is to define a difference between potential sets such that a distinct property exists between the two sets that you create.
To define two sets where unlikeness exists when they are compared.
Your job is not to merely compare three elements.
It is to compare three possible pairs of sets that can be made out of three elements.
fonix232@fedia.io 18 hours ago
And that's literally what they did.
There's a set of shapes that are filled, and a distinct set of one that is outline only.
There's a set of shapes that have 4 sides, and a distinct set of one that is 3 sides only.
There's a set of shapes that are red, and a distinct set of one shape that is green.
sp3ctr4l@lemmy.dbzer0.com 18 hours ago
If you pick A, B is also red, and C is also an irregular 4-gon. So A is not unlike either B or C.
If you pick B, A is also red, and C is also filled solid with color. So B is not unlike either B or C.
But if you pick C, while C does have elements in common with A and B…
(it shares ‘irregular 4-gon’ with A, and ‘solid color fill’ with B)
… it is also unlike each of them singly, as well as both of them together, in that it is green.
Only when you pick C do you result in a pair of sets that are cleanly dvided by the same property difference.
Is that more clear?
If you pick C, the distinction between C and A is the same distinction between C and B.
Thus, if you pick C, C is unlike A and B in the same way.
This is what I would call a clean or clear distinction, or … kind of unlikeness.
This is not the case, does not occur, if you pick A or B.
You end up with a picked set of one element that differs from the remainder set in ways that are inconsistent among the elements of the remainder set.
IE, a muddled or inconsistent distinction.
KaChilde@sh.itjust.works 18 hours ago
… only one choice is a triangle.
How is this difficult, other than if you are shape blind?
The correct choice is B.
If you pick A, B is also red, and C is also an irregular 4-gon. So A is not unlike either B or C.
If you pick C, A is also an irregular 4-gon, and B is also filled solid with color. So C is not unlike either A or B.
But if you pick B, while B does have elements in common with A and C…
(it shares ‘red’ with A, and ‘solid color fill’ with C)
… it is also unlike each of them singly, as well as both of them together, in that it is a triangle.
B is the only choice where ‘is unlike the other two’… is true, in any sense.
It has a distinct property, not found in any member of the remainder set, nor shared by the remainder set as a group.
KaChilde@sh.itjust.works 18 hours ago
What a fool you are!
… only one choice is an outline.
How is this difficult, other than if you are line blind?
The correct choice is A.
If you pick B, A is also red, and C is also a filled solid. So B is not unlike either A or C.
If you pick C, A is also an irregular 4-gon, and B is also filled solid with color. So C is not unlike either A or B.
But if you pick A, while A does have elements in common with B and C…
(it shares ‘red’ with B, and ‘4-gon’ with C)
… it is also unlike each of them singly, as well as both of them together, in that it is a triangle.
A is the only choice where ‘is unlike the other two’… is true, in any sense.
It has a distinct property, not found in any member of the remainder set, nor shared by the remainder set as a group.
sp3ctr4l@lemmy.dbzer0.com 18 hours ago
Welp.
I tap out, you’re right lol.
Don’t attempt set theory before breakfast, otherwise you end up making a fool of yourself as I have.
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