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C=-1はゼロ除算で不定で解無しだから
それ以外で定義すると約分が可能になる。
1=4
解無しだなｗ
Total BS
translated by deep L : Why make things so complicated when they can be so simple? And why make us waste our time? I can’t even believe Harvard’s involved.
A TV quiz show, perhaps?
Method 3: for every single value in number set R (real numbers) we know
c = c <->
c + 1 = c + 1 <->
c + 1 = 1 + c < – >
(c + 1)/(1 + c) = 1 <–>
(c + 1)/(1 + c) ≠ 4.
No solutions.
C=-1
c+1 devided by c+1 is allways 1, which never can be zero. You need two seconds to realize this.
There are a lot of unnecessary steps in the video for this trivial problem.
Considering addition is commutative, c+1 = 1+c. The fraction is identically 1 for all values of c (with arguably the exception of c = -1; even there, the limit is symmetrically 1 whether approached from the left or the right). Why waste 6 minutes squaring things up? Never mind Harvard, you wouldn't get into the local community college. Disliked and asked for the channel not to be recommended.
What kind of nonsense is this? Harvard entrance exam my shiny metal a**…
(c+1) is absolutely the same thing as (1+c) so far I know, so this is not even an equation as all you get is 1 = 4, which is untrue, so the problem is not solvable.
good luck in harvard
This solution (which is not a solution, but a totally useless overcomplication) would fail you in high school.
1. Assume 1+c = a
2. Let's state the obvious that 1+c = c+1 under all circumstances
3. let's state that 1+c <> 0 therefore -1 is excluded
4. declare that a/a=4 is not possible,
5. realize it's short even for a short, no youtube money
6. move over to another science
This must be a joke. Whatever c is, the fraction reduces to 1/1, which is 1. The only exception would be if c = -1, which is invalid as we can't divide by 0.
You cannot even enter a community college.
"For the fraction to be defined, c≠-1.
Then the fraction has the same numerator and denominator,
and they can be reduced. The fraction becomes 1 and it cannot be 4."
When you multiply both sides by (1 + C) you have to write down, that you can do it only if C is not equeal to -1. Therefore this whole computation is just wrong.
In method 1, why would you go the route to square both sides? That is not a logical step.