Today’s class was a relief. I figured out what an image set actually is, why image sets are not always equal to the codomain, why 4 = 2, and what happens when A and B are different sizes! I’m feeling pretty nice about it right now.

#### What An Image Set Is..Actually/Why B≠Im(f)

An image set is the set of outputs that a function can equal. The codomain may or may not be equal the image set. For example, f: ℝ→ℝ, where f = x². The image set of f [or Im(f)], is all reals greater than or equal to zero. The codomain is bigger than the image set, only because x² can never spit out negative numbers. Therefore, this function is not onto. BUT. If you change the codomain to ℝ≥0, the set is onto, because the image set and codomain are equal. When these two are equal, it means that every possible output that a given function could spit out is “accounted for” in a sense by the codomain.

Let’s try one more example to see if I got it. f: {1,2,3} → {5,6,11}, where f = 3x+2. The image set can be found by plugging in the inputs, and seeing if any of the outputs match up with anything in the codomain. The outputs for plugging in 1 and 3 are included in the codomain, so the Im(f) = {1,3}. This means that this function is also not onto, because with the inputs available (1,2 and 3), you can’t get the output of 6. It’s possible to get 6 by plugging in 4/3, but that is not an input choice in the domain. Therefore, all of the elements in the codomain are *not* accounted for.

#### 4 = 2

Casey started out class by proving this to us. Take a triangle with sides 1, 1, and √2 (by pythagorean theorem). The hypotenuse is √2, so if you were to drive a car along that length, you would go for √2. But if you decided to go in a criss-cross manner, starting at the top corner of the triangle, going down .5 units, then traveling over .5 to touch the hypotenuse, then down .5 units, then over (touching the bottom of the triangle) .5 units again to get to the final destination of the bottom corner of the triangle.

Drawing this would be way more beneficial, but blogs suck at that feature, so just go with it. Adding up all the .5 distances, you go for 2 units to get to the same point as you would have if you had just driven on the hypotenuse. Do it again with more turns, and only going .25 before each turn, it’s a more confusing route, but you still go for 2 units. When you take the limit of these lengths as the number of turns you make approaches infinity, you get 2.

BUT, when you look at the line that starts approaching, it’s √2. Therefore, √2 = 2, making 4 = 2.

When you think about it, if you’re taking a limit, there will always be really small triangles with all of their hypotenuses touching that larger hypotenuse of √2. If you find the hypotenuse of each of the .5 triangles, each one has a √2/2 hypotenuse, which, multiplied by two, equals √2. When you find the hypotenuse of a triangle with 1/4 side lengths, the hypotenuse is √2/4, and four of those hypotenuses equals: √2.

I think that’s starting to put a hole in Casey’s argument that 2 = 4. The limit of the drunken turning path may be two, but that drunken path’s hypotenuse sum will always equal √2.

#### When sets are different sizes.

We started on a worksheet in class today, and working over the first problem, I figured something out. If the domain is smaller than the codomain, out of the many possible functions between the two sets, none of them will be onto, because you will always have elements from the codomain left over (at least if you want that equation to pass the horizontal line test). If the domain is larger than the codomain, out of all the possible functions between the two sets, none of them will be one-to-one, because each input must go to some output, forcing inputs to double, tripe, or more-up on outputs.

So boom, that’s my conjecture.

Happy Monday!

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