I will prove that one is equal to two.
Firstly, a = b - Step 1.
Second, a 2 = a b - Step 2.
Third, a 2 − b 2 = a b − b 2 - Step 3.
Next, ( a + b ) ( a − b ) = b ( a − b ) - Step 4.
Then, a + b = b - Step 5.
Sixth, by a = b , then b + b = b - Step 6.
Seventhly, 2 b = b - Step 7.
FInally, ∴ 2 = 1 - Step 8.
What step did I make the first mistake? (Please enter a number only.)
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And if dividing by zero is allowed in mathematics, then really 1=2 proofs are many exist.
And, I can prove that 2 = 3 .
Firstly, a = b .
Next, a 3 = a 2 b .
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 2 + a b + b 2 = b ( a + b ) .
By a = b , b 2 + b 2 + b 2 = b 2 + b 2 .
3 b 2 = 2 b 2 .
∴ 3 = 2 .
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By the first hypothesis, a = b , then 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.