August 20, 2014
Hiroshi Michiwaki
Proof that argument for irrationality of division by zero observed in multiplication as
an inverse operation of division is incorrect
Because the relationshipholds, where a, b, c ∈ R, in a method (proof) that shows the irrationality of division by zero,
when b = 0 where a ≠ 0, we obtain a = 0, which is irrational. Therefore, division by zero, ora/0, is not defined. This specifically states the irrationality of:(Here, the result of 0 × c seen on the right side is 0. Note that it is easy to show that there is
no real number that does not become 0 when multiplied by 0. We will show this in the lemma
below.)
However, this argument is not only incomplete as a proof, but also incorrect. We will show
this below.
Theorem: The relationship between division and multiplication as inverse operations cannot
prove the irrationality of division by zero.
Proof: For the relationship in Equation (1) to hold, the following assumption is necessary. In
other words, whether the relationship in Equation (1) can hold depends on the following.
Both sides of Equation (3) are multiplied by b:

Which results in

The left side of Equation (4) becomesIn other words, it is a question of whether

generally holds or not, meaning the derivation of irrationality of division by zero in therelationship of Equation (1) above assumes that 0/0 = 1 holds.
However, whether 0/0 = 1 holds or not must be proven. Therefore, whether Equation (1)generally holds or not depends on whether 0/0 = 1 holds or not. Thus, the above irrationality
occurring in the relationship of Equation (1) alone cannot deny division by zero.
Thus, if we assume 0/0 = 1, and A = B ∈ R, and multiply both sides with B
AB = B2(7)and then subtract A2 from both sides
AB – A2 = B2 – A2(8)This is rearranged to obtainA(A – B) = (A + B)(A – B)

(9)Then, the assumption clearly allows for division of both sides by (A – B), and because that
leads toassuming A = B, it is immediately reduced toTherefore, because A is an arbitrary real number, if A ≠ 0,
1 = 2 (12)If A = 0, based on the assumption, Equation (11) becomesA = 2A (13)and if both sides of this equation are divided by 0,Therefore,
1 = 2 × 1 (15)Therefore
1 = 2 (16)Equations (12) and (16) are both irrational. The reason such irrationality occurred is because
0/0 = 1 was assumed to hold. Therefore, 0/0 = 1 does not hold.
As such, it was proven that it is impossible to show that division by zero is irrational in therelationship of Equation (1).
By this theorem, it was shown that it is impossible to show the irrationality of division byzero in the relationship of division and multiplication as its inverse operation. However, thistheorem shows that defining division as an inverse operation of multiplication is a narrow
definition of division. In other words, such a definition of division is limited to cases where
the relationship b/b = 1 holds for the divisor b in division a/b. Division with such a limiting
condition is indeed a narrow definition of division.
Lemma: There is no real number which does not become 0 (real number 0) when multipliedby 0 (real number 0).
Proof: Let us assume that there is a special value 0 that does not become 0 if multiplied by 0,
but can be applied for general numbers in processes such as the four basic arithmetic
operations, distributive property, and associative law. At this time, if we multiply both sidesof0=0+0
By 0, we obtain0 × 0 = (0 + 0) × 0=0×0+0×0
Here, if we use notation 0 × 0 = μ ≠ 0, the above equation becomes
μ = 2μThus,

=0
is obtained. However, this is inconsistent with the assumption. In other words, for any real
number x, the solution of multiplying by 0 is 0; meaning, it must be
x×0=0
and a = 0 is no exception. Of course, in the case of 0 × 0, it must be
0×0=0