Late Sunday night poser
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10-26-2024, 11:09 AM
(This post was last modified: 10-26-2024 11:18 AM by teenix.)
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Late Sunday night poser
Why are there so many problems published like...
2 divided by 2 divided by 2 divided by 2 = ? If you solved it as written, like using an HP-35, the answer would be displayed as .25 (1/4) The question itself is totally ambiguous and has no definitive answer, because like a politician, the answer can be twisted around as they like, so what is the purpose except confusion. I could say 2 divided by 0 equals 2. Wrong you say? My reasoning is 2 divided by nothing means the 2 was divided by something that doesn't exist and therefore never changed. Is that true, who decides? Late Sunday night rant :-) |
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10-26-2024, 11:39 AM
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RE: Late Sunday night poser
Saturday night? OK
But it's too heavy for a Sunday night... --Bob Prosperi |
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10-26-2024, 11:47 AM
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RE: Late Sunday night poser | |||
10-26-2024, 11:53 AM
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10-26-2024, 11:54 AM
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RE: Late Sunday night poser
(10-26-2024 11:53 AM)teenix Wrote:(10-26-2024 11:39 AM)rprosperi Wrote: Saturday night?Oops, retirement, time doesn't exist :-) I'm really looking forward to it, where every day is Saturday... --Bob Prosperi |
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10-26-2024, 11:57 AM
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RE: Late Sunday night poser
(10-26-2024 11:09 AM)teenix Wrote: The question itself is totally ambiguous and has no definitive answer From Associative property: Left-associative operations include the following: Subtraction and division of real numbers \( x-y-z = (x-y)-z \) \( x/y/z = (x/y)/z \) Right-associative operations include the following: Exponentiation of real numbers in superscript notation \( x^{y^{z}} = x^{(y^{z})} \) Conclusion: \( 2/2/2/2 = ((2/2)/2)/2 = \frac{1}{4} \) (10-26-2024 11:09 AM)teenix Wrote: I could say 2 divided by 0 equals 2. Wrong you say? Usually \(\frac{a}{b}\) is defined as the solution \(x\) of the following equation: \( a = b \cdot x \) In your case that would be: \( 2 = 0 \cdot x = 0 \) This leads to a contradiction. Therefore, there's no such \(x\). |
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10-26-2024, 05:57 PM
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RE: Late Sunday night poser
(10-26-2024 11:09 AM)teenix Wrote: Why are there so many problems published like... This reminds me of the entire 6/1(1+2) mess, which can be mitigated with a small amount of clarification. |
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10-26-2024, 08:13 PM
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RE: Late Sunday night poser
(10-26-2024 11:09 AM)teenix Wrote: Why are there so many problems published like... Since division is performed left to right, according to PEMDAS, you should do it this way: ((2 divided by 2) divided by 2) divided by 2 ((2 / 2) / 2) / 2 (1 / 2) / 2 .5 / 2 .25 There's no ambiguity at all. Where did you get 2 divided by nothing? Tom L Cui bono? |
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10-26-2024, 08:21 PM
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RE: Late Sunday night poser
(10-26-2024 05:57 PM)Eddie W. Shore Wrote: This reminds me of the entire 6/1(1+2) mess, which can be mitigated with a small amount of clarification. Every red-blooded American (and HP Prime) knows that's 18. 6 / 1 * (1 + 2) (6 /1) * (3) 6 * 3 18 Tom L Cui bono? |
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10-26-2024, 08:25 PM
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RE: Late Sunday night poser
(10-26-2024 08:13 PM)toml_12953 Wrote:(10-26-2024 11:09 AM)teenix Wrote: Why are there so many problems published like... Of course there's no ambiguity: it's clearly 2/(2/(2/2)) = 1 and anything else is plainly wrong. ;-) |
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10-26-2024, 09:39 PM
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RE: Late Sunday night poser
(10-26-2024 08:13 PM)toml_12953 Wrote: Since division is performed left to right, according to PEMDAS As far as I know, PEMDAS is about operator precedence. I guess that left associativity of subtraction is just assumed as given when chain calculations are introduced in primary school. But given that both addition and subtraction are binary operation, an expression like \(8 - 3 + 5\) doesn't even make sense. You'd have to use parentheses to either mean \((8 - 3) + 5 = 10\) or \(8 - (3 + 5) = 0\). Defining subtraction to be left associative, allows to write expressions like these without parentheses: \( \frac{1}{1+x}=1-x+x^2-x^3+x^4-x^5+ \cdots \; (|x|<1) \) |
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10-27-2024, 02:28 AM
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RE: Late Sunday night poser
(10-26-2024 09:39 PM)Thomas Klemm Wrote:(10-26-2024 08:13 PM)toml_12953 Wrote: Since division is performed left to right, according to PEMDAS Yes, it is about precedence, but the complete rule states that operators of equal precedence are to be performed left to right. Addition and subtraction are of equal precedence so they're to be performed left to right. Likewise for multiplication and division. Performing left to right isn't assumed, it's explicitly stated in the full definition (at least in the USA) Also note that implicit and explicit multiplication are equivalent (again, in the USA) Implied multiplication has no precedence over multiplication where the sign is written. Tom L Cui bono? |
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10-27-2024, 04:10 PM
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RE: Late Sunday night poser
(10-27-2024 02:28 AM)toml_12953 Wrote: but the complete rule states There are no complete rules - PEMDAS is just a mnemonic. I've never seen the term 'implied multiplication' in any textbook or book of algebra. I believe it originated on a blog. Instead, they have been 'terms' or 'products', which are made up of coefficients and factors. PEMDAS applies to operators working on terms. 1÷2a means 1÷(2 x a), because 2a is a term. It has already been multiplied. And there is historical evidence to show that this is how it has been done. I was trying to find the historical references, but it looks like internet archive is still down, though this video* has some historical references. Thankfully, with RPN, at least our calculators get it right for both of us. *I don't completely agree with that video - she talks about the books 'breaking the rules' with their own examples, but you only think rules have been broken if you don't spot the terms in the expressions. |
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10-27-2024, 05:43 PM
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RE: Late Sunday night poser
(10-27-2024 04:10 PM)dm319 Wrote:(10-27-2024 02:28 AM)toml_12953 Wrote: but the complete rule states In the USA, some teachers say that 1÷2a is the same as 1 ÷ (2 * a) since the obelus (÷) groups everything on the left as a unit and everything on the right as a unit. Others say that multiplication and division are performed left to right so 1÷2a = (1 ÷ 2) * a. Go figure! Tom L Cui bono? |
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10-27-2024, 06:17 PM
(This post was last modified: 10-27-2024 06:22 PM by Steve Simpkin.)
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RE: Late Sunday night poser
(10-27-2024 05:43 PM)toml_12953 Wrote: ... Indeed.I find it interesting that several calculator manufacturers have changed their approach to the priority of "implied" multiplication, for example 6÷2(1+2), over time. TI KB: Solution 11773: Implied Multiplication Versus Explicit Multiplication on TI Graphing Calculators. "Does implied multiplication and explicit multiplication have the same precedence on TI graphing calculators? Implied multiplication has a higher priority than explicit multiplication to allow users to enter expressions, in the same manner as they would be written. For example, the TI-80, TI-81, TI-82, and TI-85 evaluate 1/2X as 1/(2*X), while other products may evaluate the same expression as 1/2*X from left to right. Without this feature, it would be necessary to group 2X in parentheses, something that is typically not done when writing the expression on paper. This order of precedence was changed for the TI-83 family, TI-84 Plus family, TI-89 family, TI-92 Plus, Voyage™ 200 and the TI-Nspire™ Handheld in TI-84 Plus Mode. Implied and explicit multiplication is given the same priority. Please see the graphing calculator guidebooks for additional information." See the following Internet Archive page for a snapshot of this TI KB page. https://web.archive.org/web/201911051758...tId=103110 Casio has also changed their priority of "implied" multiplication over time or for different market areas. |
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10-27-2024, 07:29 PM
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RE: Late Sunday night poser
(10-27-2024 05:43 PM)toml_12953 Wrote: In the USA, some teachers say that 1÷2a is the same as 1 ÷ (2 * a) since the obelus (÷) groups everything on the left as a unit and everything on the right as a unit. Others say that multiplication and division are performed left to right so 1÷2a = (1 ÷ 2) * a. Go figure! We all agree that: \( \begin{align} \log(a \cdot b) &= \log(a) + \log(b) \\ \\ \log(a \div b) &= \log(a) - \log(b) \\ \end{align} \) It makes an easy correspondence of \(\cdot\) to \(+\) and \(\div\) to \(-\). We also know that both \(\cdot\) and \(+\) are associative operations. So we don't need parentheses: \( \begin{align} a \cdot b \cdot c &= (a \cdot b) \cdot c \\ &= a \cdot (b \cdot c) \\ \\ a + b + c &= (a + b) + c \\ &= a + (b + c) \\ \end{align} \) And we might agree that both \(\div\) and \(-\) are left-associative operations. So we don't need parentheses either: \( \begin{align} a \div b \cdot c &= (a \div b) \cdot c \\ \\ a - b + c &= (a - b) + c \\ \end{align} \) But here comes the evil math teacher with an arbitrary rule as an exception: \( a \div b c = a \div (b \cdot c) \) This leads to: \( \log(a \div b c) = \log(a) - \left(\log(b) + \log(c)\right) \) And out of thin air parentheses appear. That'll teach them. Seriously, don't use that notation. Either write: \( \frac{a}{b} c \) Or then: \( \frac{a}{bc} \) |
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11-06-2024, 06:12 PM
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RE: Late Sunday night poser
In my simplistic mind:
2 divided by nothing is equivalent to 2 is not divided by anything. 2 divided by zero is a formula for people like Bertrand Russell to explain. |
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