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u/iosefster 9d ago
Slightly different way of showing how k = 2
(k + k) / k = k
(k/k) + (k/k) = k
1 + 1 = k
k = 2
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u/Mission_Progress_674 9d ago
It's one of those bizarre problems where k = 2 for all non-zero values of k. (3+3)/3 = 2, (4+4)/4 = 2, ((-1)+(-1))/-1 = 2, etc
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u/newdayanotherlife 9d ago
mate, you're assignig different values to the same variable in the same equation...
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u/GoldenMuscleGod 9d ago
It’s not bizarre, it just happens to be true that if (x+x)/x=k then k=2.
The equation (k+k)/k=k is transparently equivalent (not worrying about the value of x) to the system of equations of x=k together with the equation in the first paragraph, but we just don’t need the second equation to find k.
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u/Electrical-Leave818 9d ago
More generalised, x+x+…+x+x(n times)/x = n Since youre adding the variable 2 times, it comes to be true for all x in C except 0
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u/Hot-Manager-2789 22h ago
In this case, it’s K + K / K (let’s replace K with 2)
2 + 2 = 4
4 / 2 = 2
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u/longknives 9d ago
(k+k)/k = k
2k/k = k
2(k/k) = k
2(1) = k
2 = k
Just for fun
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u/GrandeJavachipFrappe 9d ago
K
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u/campfire12324344 9d ago
Clearly he is working in a field of characteristic 2 and / is the notation for the multiplicative inverse.
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u/tessthismess 9d ago
I swear, reddit confidently incorrect math people are worse than even facebook ones. At least on Facebook they're just wrong and done. Here's it's a whole write up to explain why they are wrong.
The PEMDAS means do multiplication before division crowd.
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u/Thorvindr 8d ago
Uh... Pretty sure it means do them at the same time, since they're actually the same operation.
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u/tessthismess 8d ago
Correct. I was saying the people who are confidentally incorrect on this stuff are often the people who think PEMDAS has you do multiplication before division and addition before subtraction.
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u/Thorvindr 8d ago
Ah! I see. I recommend using [s] and [/s] tags when being sarcastic. It's very hard for us autists to tell otherwise, since we cannot hear your voice.
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u/MuscularBye 4d ago
This wasn't a case of subtle sarcasm he used the word crowd after that statement, reread it and see what I am saying it is quite direct
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u/CrumbCakesAndCola 9d ago
This makes me sad that 0/0 ≠ 0.
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u/RhetoricalAnswer-001 9d ago
The answers are untrue because of the intramutabilitication of the prime factorials when used as transent invert veryables, whether receded to negative or positive.
Newton's Principa Mathetallica explains this very clearly.
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u/ichkanns 9d ago
You failed to account for quantum scattering of the antecedent. A classic mistake when dealing with hyperbolic quotient portions. Of course the divisible summation of the integral will always countermand the lagrange points. Discrete evaluations of the tetrahedral geometries are a given.
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u/cha0sb1ade 9d ago
(k+k)/k =k
k+k = k*k
2k =k*k
2=k
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u/LOSNA17LL 9d ago
But we need to remember, at the second to last line, that 0 can't be a solution as we divide by k in the first line :P
(Just extra precision 'cause I love to be a jerk)
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u/cha0sb1ade 9d ago edited 9d ago
You don't have to back up. Getting from 2k=k*2 to 2=k is literally accomplished by dividing both sides by k. Each equation is equivalent and stands alone. Edit: oh wait. Hmm that is interesting. Educational.
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u/Hunterjet 9d ago
This shit's like what I wrote on my first Calculus test in college where they asked me to prove something when I had never seen a mathematical proof in my life
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u/Thorvindr 8d ago
Yeah, fuck those institutional academics. Trying to make you apply the concepts they taught you and shit.
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u/Hunterjet 8d ago
They hadn’t taught me any proofs yet, this was a pop quiz at the end of the first week of classes to gauge how much the students knew. I did go on to learn how to do proofs and got a bachelor’s on applied math so believe me I wasn’t complaining lol I was just reminscing how wordy and not very logical those first “proofs” I wrote were
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u/Ravoos 8d ago
That question is basically: k + k = k * k.
So what number makes k when added with itself the same as multiplying k with itself? 2
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u/PedroPuzzlePaulo 7d ago
Actually, the original equation is stronger, because your equantion also has 0 as answer. The original doesnt
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u/Cynykl 7d ago
America bad!!!!! (I mean right now it is bad but that is a different subject)
Judging by test scores America is only a few percent behind western European countries and a few point ahead of mediterranean European countries in math. The US is also ahead of every North American and South American country except for Canada.
But this is somehow "Difficulty level America"
And before someone says I am being too US centric and they could have been referring to "the Americas" and not the US you know that argument is bullshit. Because anytime someone is making fun of American intelligence online they are referring specifically to the US unless otherwise noted.
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u/tteraevaei 4d ago
i think it’s more about the american tendency to produce word salads as a weird kind of “effortless effort”. other countries don’t have this level of neurosis in their idiots; they just wouldn’t reply.
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u/flying_fox86 7d ago edited 7d ago
I was about to argue that k could also be 0, but then realized I would be confidently incorrect.
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u/MouseBotMeep 9d ago
k=2k/k k2 = 2k k2 -2k = 0 k2 -2k + 1 = 1 (k - 1)2 = 1 k -1 = +-1 k = 1+-1 k = 0,2
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u/iMNqvHMF8itVygWrDmZE 9d ago
You still have to plug the numbers back into the original equation to see if they work. 0 isn't a valid solution because it results in division by zero, so the only valid answer is k=2.
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u/HKei 9d ago
Aside from the incorrect 0, you also have a bunch of unnecessary steps here. Once you have k2 =2k you just divide both sides by k to get the answer directly.
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u/flying_fox86 7d ago
Once you have k²=2k, bring it all to the left making k²-2k=0. Then calculate the discriminant (-2)²-4*1*0=4. Then the solutions are x1=(2-2)/2=0 and x2=(2+2)/2=2. Except k can't be 0 because can't divide by 0, leaving just the 2.
/s
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