Proof that 1 = 2 1 = 2 .

Algebra Level 2

I will prove that one is equal to two.

Firstly, a = b a = b - Step 1.

Second, a 2 = a b a ^ { 2 } = ab - Step 2.

Third, a 2 b 2 = a b b 2 a ^ { 2 } - b ^ { 2 } = ab - b ^ { 2 } - Step 3.

Next, ( a + b ) ( a b ) = b ( a b ) ( a + b ) ( a - b ) = b ( a - b ) - Step 4.

Then, a + b = b a + b = b - Step 5.

Sixth, by a = b a = b , then b + b = b b + b = b - Step 6.

Seventhly, 2 b = b 2b = b - Step 7.

FInally, 2 = 1 \therefore 2 = 1 - Step 8.

What step did I make the first mistake? (Please enter a number only.)


The answer is 5.

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1 solution

. .
Apr 17, 2021

By the first hypothesis, a = b a = b , then a b = 0 a - b = 0 .

Then, in Step 5., I divided a number by zero.

Because dividing by zero is undefined, so I did make the first mistake in Step 5.

And if dividing by zero is allowed in mathematics, then really 1=2 proofs are many exist.

And, I can prove that 2 = 3 2 = 3 .

Firstly, a = b a = b .

Next, a 3 = a 2 b a ^ { 3 } = a ^ { 2 }b .

a 3 b 3 = a 2 b b 3 a ^ { 3 } - b ^ { 3 } = a ^ { 2 }b - b ^ { 3 } .

( a b ) ( a 2 + a b + b 2 ) = b ( a 2 b 2 ) ( a b ) ( a 2 + a b + b 2 ) = b ( a + b ) ( a b ) ( a - b ) ( a ^ { 2 } + ab + b ^ { 2 } ) = b ( a ^ { 2 } - b ^ { 2 } ) \rightarrow ( a - b ) ( a ^ { 2 } + ab + b ^ { 2 } ) = b ( a + b ) ( a - b ) .

a 2 + a b + b 2 = b ( a + b ) a ^ { 2 } + ab + b ^ { 2 } = b ( a + b ) .

By a = b a = b , b 2 + b 2 + b 2 = b 2 + b 2 b ^ { 2 } + b ^ { 2 } + b ^ { 2 } = b ^ { 2 } + b ^ { 2 } .

3 b 2 = 2 b 2 3 b ^ { 2 } = 2 b ^ { 2 } .

3 = 2 \therefore 3 = 2 .

. . - 1 month, 3 weeks ago

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