• BlueMagma@sh.itjust.works
      5·
      1 year ago

      Brilliant, now I wonder what ages this works for, I figured only 1 and 2, but then I realised we could write the father’s age in other bases…

      1 = 2^0 (20 b10)

      2 = 2^1 (21 b10)

      3 = 3^1 (31 b7 = 22)

      6 = 6^1 (61 b4 = 25) if they are lucky the grand father will be 61 that year :-D

      8 = 2^3 (23 b12 =27)

      9 = 9^1 (91 b3 = 28)

      14 = 14^1 (141 b4 = 33)

      • aDogCalledSpot@lemmy.zipEnglish
        1·
        1 year ago

        You have mistakes in a few of those. The number “61” doesnt exist in b4. 25b10 in b4 is “121”.

        Similar problem with 91b3 and 141b4.

  • blind3rdeye
    714·
    1 year ago

    Ah yes. How fitting for a young new person in the world. A reminder that 2°C of warming above the pre-industrial mean would be catastrophic, but also is a good lower-limit of what to expect based on current intentions.

    • milicent_bystandr
      193·
      1 year ago

      What?! Impossible to start a family at 18 and also enjoy mathematics?

      Not everyone who has unprotected sex at 18 (or with an 18 yr old) is some numbskull just going at it for unscrupulous pleasure.

      (As another reply also pointed out: the pun was crafted by the OP’s dad, not the 1yr-old’s dad; and OP could be the child’s mum or dad)

    • gun@lemmy.ml
      15·
      1 year ago

      He probably didn’t. Her dad (the grandpa) made the balloons.

  • Doctor xNo@r.nfEnglish
    221·
    1 year ago

    Damn, that took me waaay too long to get…

    Not my brightest moment… 😅

  • ThatWeirdGuy1001@lemmy.world
    233·
    1 year ago

    I know I’m bad at math but I don’t understand how 2x0=0 but 2^0=1

    How are they different answers when they’re both essentially multiplying 2 by zero?

    Someone with a bigger brain please explain this

    Edit: I greatly appreciate all the explanations but all they’ve done is solidify the fact that I’ll never be good at math 😭

    • jendrik@discuss.tchncs.de
      313·
      1 year ago

      subtracting one from Exponent means halving (when the base is two):

      2⁴ = 16 2³ = 8 2² = 4 2¹ = 2 2⁰ = 1

      It’s a simple continuation of the pattern and required for mathemarical rules to work.

      • uberrice@feddit.de
        11·
        1 year ago

        This is confidently wrong.

        3^0 is also 1. 2738394728^0 is also 1.

        Edit: just saw that technically you’re correct - sure.

        IF base 2, Exponent reduction equals to halving - dividing by 2.

        For x^y reducing y by one is equal to dividing by x, then we have the proof it always works.

        • Globulart@lemmy.world
          4·
          1 year ago

          But that’s because for 3 the sequence is dividing by 3 not 2.

          81, 27, 9, 3, 1, 1/3, 1/9, etc.

          3^4, 3^3, 3^2, 3^1, 3^0, 3^(-1), 3^(-2), etc.

          You’re not always halving, but the method is the same and it sometimes helps people understand the concept more easily.

    • Globulart@lemmy.world
      15·
      1 year ago

      This isn’t strictly speaking a proof, but it did help me to accept it as it demonstrates the function that makes it 1.

      2^3 = 2x2x2

      2^2 = 2x2

      (23)/(22) = (2x2x2)/(2x2) = 2

      = 2^(3-2)

      In general terms:

      (xa)/(xb) = x^(a-b)

      If a and b are the same number this is x^0 and obviously (xa)/(xa) is one because anything divided by itself is 1.

      Hope that helps

        • Globulart@lemmy.world
          3·
          1 year ago

          2^(a-b) = (2a)/(2b)

          You can see this in the example above but perhaps it’s better to use different powers to make things a bit clearer.

          2^5=2x2x2x2x2

          2^3=2x2x2

          (25)/(23)=(2x2x2x2x2)/(2x2x2)

          You can cancel 3 of the 2s from the top and bottom of the fraction to be left with 2x2, or 2^2.

          I.e. (25)/(23)=2^2

          The quicker way to calculate this is doing 2^(5-3) which when you resolve the bracket is obviously just 2^2 or 2x2.

          If both numbers in the bracket are the same the bracket will always resolve to 0, which is the same as saying a number divided by itself, any number divided by itself is one so it follows that any number to the power 0 is also 1 (because it’s essentially exactly the same calculation).

        • Flumsy@feddit.de
          31·
          1 year ago

          That was pretty complicated, here is a simpler answer I hsve come up with:

          1=(2x2x2)/(2x2x2)=2³/2³=2³⁻³＝2⁰

          If that makes sense to you…

    • DSTGU
      15·
      1 year ago

      0 is the neutral element for addition. This is why when we have a number then 0 + number = number (0 doesnt change the value in addition) and why 0 x number = 0 (if you add a number 0 times you will have 0). (Multiplication is adding one of the numbers to itself the number of times designated by the second number)

      The same way 1 is the neutral element for multiplication. This is why when you have some number then 1 * number = number. This is also why number^0 = 1 (if you never multiply by a number you are left with the neutral element. It would be weird if powering by 0 left you with 0 for example because of how negative powers work)

      This is the level 1 answer.

      The level 0 answer is that it is this way because all of mathematics is a construct designed to ease problem solving and all people collectively agreed that doing it this way is way more useful (because it is)

      Choose which one you want

    • TokyoMonsterTrucker@lemmy.dbzer0.comEnglish
      15·
      1 year ago

      Easiest explanation I can think of using the division law for exponents:

      Since we can use any number for the initial fraction, as long as the denominator is the same as the numerator, any number to the zeroth power is equal to 1. In general terms, then, for any number, x:

    • lugal@lemmy.ml
      8·
      1 year ago

      You can think of 1 as the “empty product” (or the “neutral element of multiplication” if you want to be fancy). 2^x means you have x factors of 2. If you have 0 factors, you have the “empty product”

    • ShaunaTheDead@kbin.social
      8·
      1 year ago

      I see other people have posted good explanations, but I think the simplest explanation has to do with how you break down numbers. Lets take a number, say, 124. We can rewrite it as 100 + 20 + 4 and we can rewrite that as 1 * 10^2 + 2 * 10^1 + 4 * 10^0 and I think you can see why anything raised to the 0th power has to equal 1. Numbers and math wouldn’t work if it didn’t.

    • Floey
      7·
      1 year ago

      I like to think of it this way:
      2^3 is the same as 2 x 2 x 2.
      But you can arbitrarily multiply by as many 1s as you want because 1 has the identity property for multiplication.
      So we can also write 2^3 as 2 x 2 x 2 x 1 x 1.
      2^2 as 2 x 2 x 1 x 1.
      2^1 as 2 x 1 x 1.
      2^0 as 1 x 1 or just 1.

      Multiplying a number by another number is the same as adding a number to itself that many times. And 0 is has the identity property for addition, so similarly:
      2 x 3 = 2 + 2 + 2 + 0 + 0
      2 x 2 = 2 + 2 + 0 + 0
      2 x 1 = 2 + 0 + 0
      2 x 0 = 0 + 0

    • LordOfTheChia@lemmy.world
      4·
      1 year ago

      In addition to the explanation others have mentioned, here it is in graph form. See the where the graph of 2^x intersects the y axis (when x=0):

      https://people.richland.edu/james/lecture/m116/logs/exponential.html

      This also has some additional verbal explanations:

      http://scienceline.ucsb.edu/getkey.php?key=2626

      The simplest way I think of it is by the properties of exponentials:

      2^3 / 2^2 = (2 * 2 * 2) / (2 * 2) = 2 = 2^(3-2)

      Dividing two exponentials with the same base (in this case 2) is the same as that same base (2) to the power of the difference between the exponent in the numerator minus the exponent in the denominator (3 and 2 in this case).

      Now lets make both exponents the same:

      2^3 / 2^3 = 8/8 = 1

      2^3 / 2^3 = 2^(3-3) = 2^0 = 1

    • GigglyBobble@kbin.social
      21·
      1 year ago

      2^0 isn’t multiplying by zero. Considering this law: 2^a / 2^b = 2^(a-b)
      it’s obvious why 2^0 = 1
      If a=b you’re dividing by the same number resulting in 1.

      Unfortunately, I cannot explain/prove the first law though.

      • zalgotext@sh.itjust.works
        3·
        1 year ago

        The first rule is just simple division:

        (2x2x2x2) / (2x2) =

        (2/2) * (2/2) * 2 * 2=

        1 * 1 * 2 * 2 =

        2 * 2 =

        4

        Writing in terms of powers:

        (2^4) / (2^2) =

        (2^(4-2)) =

        (2^2) =

        4

        The two bottom 2’s “cancel out” (really they just divide into one another to make 1’s) two of the top 2’s and you’re left with two top twos.

    • BlazingFlames6073@lemdro.idEnglish
      1·
      1 year ago

      Thanks, I couldn’t even tell what the image was about math. I thought a dirty joke was hidden somewhere involving the 0. Didn’t realize it was small and floating above on the right so people would immediately realize it’s a power lol. Many people hide clever things but I always approach them in the wrong way lol.

  • affiliate@lemmy.world
    131·
    1 year ago

    for anyone curious, here’s a “constructive” explanation of why a0 = 1. i’ll also include a “constructive” explanation of why rational exponents are defined the way they are.

    anyways, the equality a0 = 1 is a consequence of the relation

    am+1 = am • a.

    to make things a bit simpler, let’s say a=2. then we want to make sense of the formula

    2m+1 = 2m • 2

    this makes a bit more sense when written out in words: it’s saying that if we multiply 2 by itself m+1 times, that’s the same as first multiplying 2 by itself m times, then multiplying that by 2. for example: 23 = 22 • 2, since these are just two different ways of writing 2 • 2 • 2.

    setting 20 is then what we have to do for the formula to make sense when m = 0. this is because the formula becomes

    20+1 = 20 • 21.

    because 20+1 = 2 and 21 = 2, we can divide both sides by 2 and get 1 = 20.

    fractional exponents are admittedly more complicated, but here’s a (more handwavey) explanation of them. they’re basically a result of the formula

    (am)n = am•n

    which is true when m and n are whole numbers. it’s a bit more difficult to give a proper explanation as to why the above formula is true, but maybe an example would be more helpful anyways. if m=2 and n=3, it’s basically saying

    (a2)3 = (a • a)3 = (a • a) • (a • a) • (a • a) = a2•3.

    it’s worth noting that the general case (when m and n are any whole numbers) can be treated in the same way, it’s just that the notation becomes clunkier and less transparent.

    anyways, we want to define fractional exponents so that the formula

    (ar)s = ar • as

    is true when r and s are fractional numbers. we can start out by defining the “simple” fractional exponents of the form a1/n, where n is a whole number. since n/n = 1, we’re then forced to define a1/n so that

    a = a1/n•n = (a1/n)n.

    what does this mean? let’s consider n = 2. then we have to define a1/2 so that (a1/2)2 = a. this means that a1/2 is the square root of a. similarly, this means that a1/n is the n-th root of a.

    how do we use this to define arbitrary fractional exponents? we again do it with the formula in mind! we can then just define

    am/n = (a1/n)m.

    the expression a1/n makes sense because we’ve already defined it, and the expression (a1/n)m makes sense because we’ve already defined what it means to take exponents by whole numbers. in words, this means that am/n is the n-th square root of a, multiplied by itself m times.

    i think this kind of explanation can be helpful because they show why exponents are defined in certain ways: we’re really just defining fractional exponents so that they behave the same way as whole number exponents. this makes it easier to remember the definitions, and it also makes it easier to work with them since you can in practice treat them in the “same way” you treat whole number exponents.

      • Ghyste@sh.itjust.works
        22·
        1 year ago

        They are either part of the “everything is a meme” bullshit, or they don’t care whether the crap they’re posting is a meme.

        Both crowds will complain that they’re not posting to other communities because they’re “dead”. Neither group sees the irony.

        In the end, these idiots are looking for upvotes and have found the community of other idiots who will upvote anything in their feed without a care of where it’s posted. This group of idiots can go back to reddit.

  • Haus@kbin.social
    11·
    1 year ago

    It’s been a while, but I think I remember this one. Lim 1/n =0 as n approaches infinity. Let x^0 be undefined. For any e>0 there exists an n such that |x^(1/n) -1| < e. If you desire x^(1/n) to be continuous at 0, you define x^0 as 1.

    E2a: since x^(1/n)>1, you can drop the abs bars. I think you can get an inequality to pick n using logs.

    • uberrice@feddit.de
      4·
      1 year ago

      Simpler: x^1 = x, x^-1 = 1/x

      x^1 * x^-1 = x^0 = x/x = 1.

      Of course, your explanation is the “correct” one - why it’s possible that x^0=1. Mine is the simple version that shows how logic checks out using algebraic rules.