Proofs that 2 +2 = 5 By Brian Simpson

     I covered the idea that the woke politically correct were rejecting mathematical objectivity, in arithmetic truths such as 2 + 2 = 4, that 2+ 2 could equal 5. Here are some more thoughts on this, with two bogus, but delightful “proofs” and the re-definition idea that I discussed.
  https://www.quora.com/How-can-we-make-2+2-5-1?utm_source=amerika.org%22

Mayur Goswami
Answered August 8, 2019

Originally Answered: Can you prove 2 + 2 = 5?
Two proofs are shown.

i)
20−20=25−25(1)(1)20−20=25−25
Expands to a multiple of 5.
5×4−5×4=5×5−5×5(2)(2)5×4−5×4=5×5−5×5
(5−5)(5−5)When you come
4×(5−5)=5×(5−5)(3)(3)4×(5−5)=5×(5−5)
4=5(4)(4)4=5
2+2=5(5)(5)2+2=5
End of proof.
Another 1,
ii)
−20=−20(6)(6)−20=−20
So
16−36=25−45(7)(7)16−36=25−45
Make each term a product
42−4×9=52−5×9(8)(8)42−4×9=52−5×9
Modified the second term on both sides.
42−2(4×92)=52−2(5×92)(8)(8)42−2(4×92)=52−2(5×92)
On both sides (92)2(92)2 Plus
42−2(4×92)+(92)2=52−2(5×92)+(92)2(9)(9)42−2(4×92)+(92)2=52−2(5×92)+(92)2
Both sides(a−b)2(a−b)2 Because it is shaped
(4−92)2=(5−92)2(10)(10)(4−92)2=(5−92)2
That means
4−92=5−92(11)(11)4−92=5−92
On both sides 9292 Add
4=5(12)(12)4=5
∴2+2=5(13)(13)∴2+2=5
End of proof.
The proof that looks right is actually a mistake.
In other words, neither breaks the basic rules in mathematics, so it is impossible 2+2=5. It looks as if has been proved. Do you know where the rules are violated?
The first one is$(3)$From$(Four)$I'm doing something that shouldn't be done. Divide by zero. In mathematics, division by zero is prohibited. So, (3)From (4)So don't proceed, so (5)I can't get to.
The second person also violates the rules, but it may be difficult to understand.
The mistakes here (10)From (11)It is done in the place to move to.
In general
a2=b2a2=b2
time,
a=bwellthena=±ba=bwellthena=±b
is. 10 From 11 I took the “+” without permission when moving to 13 I moved to the impossible formula, but in fact, only "-" is established,(11) Is
4−92=−(5−92)(11)(11)4−92=−(5−92)
And the correct equation.
These two proofs are cool if I think of them, but they come out as if they are useful or the in the world. However it was an English video, the above two false proofs and the points of the lie are explained to the exact same question.
tried to think on my own, but changed the policy after thinking 15 minutes. This choice may have been right.
It takes more than an hour to write a formula to see the video.This choice may be a major failure.”

    The first “proof” fails because it involves division by zero, which is not defined in ordinary arithmetic, hence the proof fails. But, here is another attempt which does better:
  http://www.amerika.org/politics/2-2-5/

“In the integer system, under usual addition: 2+2=4, and it will “NEVER” be the case that 2+2=5. However, in abstract algebra, it is possible to define a new operation, let’s call it #, such that the set of integers (Z, #) under the operation # is a group.
And we can define this operation to be:
For any elements a, b in Z, we have that a#b=a+b+1.
In order for (Z,#) to be a group, the following must be satisfied:
i) For all a, b, c in Z, then a#(b#c) = (a#b)#c [Associative law]
a#(b#c) = a#(b+c+1) = (a+b+c+1)+1 = a+b+c+2
Similarly,
(a#b)#c = (a+b+1)#c = (a+b+1+c)+1 = a+b+c+2,
and thus our set follows the associative law.
ii) There exists an element e in Z such that a#e = e#a = a, for all a in Z [Identity element exists]
Let e= -1, then we see that:
a#e =a#-1 = (a +(-1)) + 1 = a +(-1) +1 = a,
e#a = -1#a = ((-1)+a)+1 = (-1)+a+1 = a,
and thus our set has an identity element, -1.
iii) For any element a in Z, there exists an element b such that a#b = e [Inverses exists]
For any element a in Z, consider the element b = -a-2,
a#b = a#(-a-2) = (a+(-a-2))+1 = a+(-a)+(-2)+1 = -1,
b#a = (-a-2)#a = ((-a-2)+a)+1 = (-a)+(-2)+a+1 = -1,
Hence, we have an inverse b under our operation # for any element a in Z. Now, we have proved that the integers under this operation “#” form a group. Henceforth, we can say 2#2 = 5. And since the “+” symbol is merely just a symbol for an operation, in all theoretical purposes, 2+2 =5 can be shown if the operation “+” is defined as the above “#”.
Finally, for all intents and purposes, as I stated from the very beginning, 2+2 is NEVER equal to 5. But, to answer the original question: “How can we make 2+2=5?”, I believe this is a neat way of doing so using a different type of addition from an abstract algebra perspective.”

     This goes along the lines I suggested in a previous article to simply re-define the arithmetic operators, which no doubt under multiracial President Kamila Harris, will be standard practice.