1=2 Proof

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In this video, we explore a false proof that 1 equals 2. This proof has circulated online for years and continues to fool people with its clever manipulation of mathematics. However, upon closer inspection, it becomes clear that the proof relies on flawed logic and is not a valid demonstration of the equality of 1 and 2. Despite its convincing appearance, this proof is nothing more than a mathematical trick designed to mislead and deceive. Don't fall for it! Stay vigilant and always be sure to carefully examine the reasoning behind any mathematical claims you encounter.

#math #brithemathguy #shorts

Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.

[loatchi_le]

Traceback (most recent call last):
File "<stdin>", line 4, in <module>
ZeroDivisionError: division by zero

[mrcucarachaman]

If y’all didn’t know yet this didn’t work because when you cancel (divide both sides) by (a-b) since a = b you’re dividing by 0

[NoName-rdet]

Most of these 2=1 errors are usually the result of a hidden divide by 0 somewhere

[MoldManMotionGrafix]

I like how he walked us down this path of deception then mocked us.

[nickronca]

I saw a video very similar to this one time "proving" that 4=3 (by dividing by zero on both sides like this equation does), and the video started off with a+b=c, and someone in the comments section INSISTED over and over again that the mistake was assuming a+b=c in the first place. No matter how much I tried to point out that a+b can equal c all you want because you can set the numbers equal to whatever you want and the actual mistake was that you are dividing both sides by zero, he insisted that it must be because a+b can not equal c. No matter what logic or reasoning I tried to use to show why that can be the case if you want it to be.

[terranceparsons]

1 = 2 eh? OK I will give you $1 and you give me $2, because they are the same.

[veerusirimalla]

Since,
a=b
a-b=0

Therefore, a-b can't be cancel out at your forth step

[BeyX]

Here’s my approach
Notice that we assumed a=b
a2 = ab
a2 - b2 = ab - b2
(a+b)(a-b) = b(a-b)
But since a=b
(b+b)(b-b) = b(b-b)
(2b)(0) = b(0)
0 = 0
So the equation becomes true at the fourth step if you substitute a = b. And whatever occurred after it was a manipulation which led to 1=2.

[insertfunnynamehere]

Now using standard algebraic rules, divide both sides by 0 😂

[Pridestein]

The way I know it is that when you get to the a+b=b stage, subtract b from both sides to get a=0. Since we never specified what a was, a could be anything - therefore, ANY NUMBER EQUALS ZERO! (The proof is still false for the same reason as the one in the video.)

[mrlitty]

Ah yes, once again the classic dividing by zero

[nickcampbell]

Yeah, anytime someone trys to "cancel" something, it is good to check what they are doing.

In order to "cancel" (a-b), you would have to divide by it. This is unfortunately not supported in the current scenerio. Since a=b is the premise, then a-b=0 is a natural result and dividing by a-b would be dividing by 0.

[sungkarson]

“See you soon!” He said to his friend
“See you soon…” The old man said to his wife
They look the same, but they are not

[yairkaz]

There are 2 errors, 1 that you showed, where you divine by 0 because a-b=0. However there is another as you cannot divine b before checking it doesn't equal 0. And in the equation 2b=b b must be equal to 0 for it to exist, hence why you also can't do this.

[_bnarayanrahul]

When you were dividing with (a-b) on both sides, it must not be equal to zero as it is in the denominator, then a not equals b which contradicts the first equation.

[prodromoskonstandas]

You can't even cancel the b because it could be also equal to zero

[sanjeetkosle]

While cancelling rhs with lhs you are actually transferring rhs to lhs and dividing it which is equal to 1 then you are multiplying that 1 with remaining values, but according to bodmas rule, first solve the bracket and then divide, so (a-b) on both side will be equal to 0 resulting in equation 0=0

[elmaruchiha]

Let do the same, but with a approaching b:
a→b
=> a²→ab and a-b→0
=>a²-b²→ab-b² and a-b→0
=> (a+b)(a-b)→b(a-b) and a-b→0
Since 0 doesn't approach 0, a-b→0 cause a-b≠0.
=>(a+b)(a-b)→0 and a-b≠0
=>a+b→0
=>a→-b
So we started with a→b and ended by a→-b.
So a→b and a→-b.
Where is the mistake?

[BurningShipFractal]

From the title “1=2 Proof”, I immediately knew that this video is on dividing by zero.

[OptimusPhillip]

The thing I find fun about this "proof" is that even after the divide by zero, it can still be salvaged if you consider that a and b could be equal to 0