• 7 Posts
  • 1.27K Comments
Joined 2 years ago
Cake day: June 7th, 2023
  • saigottopolitics @lemmy.world•Four Democrats Pass Bill Making It Harder for Married Women to Vote
    17·
    4 days ago

    I live in Canada, I can vote using my free government issued healthcard or I can bring a friend to vouch for me, or i can bring a student id and a bill. While most people probably vote with their drivers license or photo ID this enables people who are homeless, very old, or in my case in 2021, just moved. (Here’s what’s needed for the curious). You’ll notice in that link there are special exemptions for people who live in long term care homes, for whom it is much more common to have no form of id.

    People who don’t have easy access to id are societies most vulnerable people and I think it is especially important that they have access to voting.

    America does not have a free form of id (in most states anyway) and does not allow someone to vouch as a form of identification.

  • saigotto196@lemmy.blahaj.zone•RuleEnglish
    23·
    5 days ago

    Here is my attempt to eli5, a metric is a formalized/generalized way to describe distance. Smart people thought about what makes distance distance and basically made a set of rules. Distance is a function where the distance between a point and itself is 0 (and only 0 in that case), is always positive, is the same distance whether you are coming or going and that going to a place and then another place has at least as much distance as just going to the last place (which is kind of the same as saying the shortest path between 2 points is a straight line).

    You can see how these rules apply to point in 3d(or 2d) space and our intuitive understanding of distance between them. For example If a store is 2km going to a bank then the store is at least 2km but maybe more and if its 2km from home to the store its also 2km from the store to home. This might seem obvious, and it is for 3d space, but we can take it and apply it to all kinds of things.

    This question is intentionally convoluted, but one way of conceptualizing it is: 🍎🍇🍌 are each functions that takes one value and spits out another. If you would graph this function it makes a line. 🍊 takes 2 lines and tells us how far apart they are, you can think about many ways to compare how far apart 2 line are, but the one given to us is to just take the one where the difference between the heights of the lines is greatest. For an example lets say 🍎 is the price of eggs and 🍇 is the price of organic eggs then 🍊(🍎,🍇) would give us the biggest difference in price there has ever been between them.

    Our task in the problem is to prove that that idea of distance given to us follows the same rules as our intuitive definition of distance.

    E: I originally misread the values the functions took as 2 dimensional coordinates, but it is really just 1 dimensional data, so I changed the metaphor.

  • saigotto196@lemmy.blahaj.zone•RuleEnglish
    4·
    5 days ago

    It should be ||🍎(x) -🍌(x)| +|🍌(x) - 🍇(x)|| >=|🍎(x) -🍌(x) +🍌(x) - 🍇(x)| = |🍎(x) - 🍇(x)| I missed the abs that I added in the previous step.

    let me make the variables less annoying:

    ||x-y|+|y-z|| >= |x-y+y-z| = |x-z| we are getting rid of the abs around |x-y| and |y-z| so the 2 y’s can cancel out. We can do this because |x-y| >= x-y because |q| >= q

  • saigotto196@lemmy.blahaj.zone•RuleEnglish
    92·
    2 days ago

    It’s been a while but here we go:

    for orange to be a metric 4 conditions must be met:

    1. 🍊(🍎,🍎) = 0
    proof

    since 🍎(x) - 🍎(x) will always be 0 for any 🍎 and any x in domain

    1. 🍊(🍎,🍌) > 0 if 🍎 != 🍌.
    proof

    |🍎(x) - 🍌(x)| >= 0 by definition, so 🍊(🍎,🍌) must be >= 0. we only have to prove that:

    🍊(🍎,🍌) = 0 -> 🍎=🍌

    Consider the contrapositive: 🍎!=🍌 -> 🍊(🍎,🍌) != 0

    since 🍎!=🍌 ∃x s.t 🍎(x) != 🍌(x)

    but then |🍎(x) - 🍌(x)| > 0

    thus 🍊(🍎,🍌) > 0

    thus 🍊(🍎,🍌) = 0 -> 🍎=🍌

    1. 🍊(🍎,🍌) = 🍊(🍌,🍎)
    proof

    |🍎(x) - 🍌(x)| = |-1(-🍎(x) + 🍌(x))|

    |-1(-🍎(x) + 🍌(x))| = |-1(🍌(x) - 🍎(x))|

    |-1(🍌(x) - 🍎(x))| = |🍌(x) - 🍎(x)| since |-q| =|q|

    so for any x |🍎(x) - 🍌(x)| = |🍌(x) - 🍎(x)|

    which means 🍊(🍎,🍌) = 🍊(🍌,🍎)

    1. The Triangle Inequality:🍊(🍎,🍇) <= 🍊(🍎,🍌) + 🍊(🍌, 🍇)
    proof

    let x be the element in [a,b] s.t |🍎(x) - 🍇(x)| is maximized

    let y be the element in [a,b] s.t |🍎(y) - 🍌(y)| is maximized

    let z be the element in [a,b] s.t |🍌(z) - 🍇(z)| is maximized

    🍊(🍎,🍇) <=🍊(🍎,🍌) + 🍊(🍌, 🍇) is equivalent to

    |🍎(y) -🍌(y)| +|🍌(z) - 🍇(z)| >= |🍎(x) - 🍇(x)|

    Let’s start with the following (obvious) inequality:

    |🍎(y) -🍌(y)| +|🍌(z) - 🍇(z)| >= |🍎(y) -🍌(y)| +|🍌(z) - 🍇(z)|

    |🍎(y) -🍌(y)| +|🍌(z) - 🍇(z)| >= |🍎(x) -🍌(x)| +|🍌(z) - 🍇(z)| since |🍎(y) - 🍌(y)| is maximized

    |🍎(x) -🍌(x)| +|🍌(z) - 🍇(z)| >= |🍎(x) -🍌(x)| +|🍌(x) - 🍇(x)| since |🍌(z) - 🍇(z)| is maximized

    |🍎(x) -🍌(x)| +|🍌(z) - 🍇(z)| >= ||🍎(x) -🍌(x)| +|🍌(x) - 🍇(x)|| since |q| + |p| >= 0 so |q| + |p| = ||q| +|p||

    ||🍎(x) -🍌(x)| +|🍌(x) - 🍇(x)|| >=|🍎(x) -🍌(x) +🍌(x) - 🍇(x)| = |🍎(x) - 🍇(x)| since |q| >= q forall q

    therefore |🍎(y) -🍌(y)| +|🍌(z) - 🍇(z)| >= |🍎(x) - 🍇(x)|

    since all 4 conditions are satisfied the 🍊 is a metric!