If 1/0 is infinity and 2/0 is infinity, why isn't 1=2?

Ramankholi said:
If 1/0 is infinity and 2/0 is infinity, why isn't 1=2?

Please Somebody help me out!
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What you have said here is exactly why we don't define division by zero, and don't treat infinity as a number. If we did, then we would get false results as you did. So we don't.
 
For one thing, infinities are not equal. For example there are more irrational numbers than there are integers. In fact there are an infinite number of different infinities.
 
Ramankholi said:
If 1/0 is infinity and 2/0 is infinity, why isn't 1=2?
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As others have pointed out, zero division is not infinity. And if it were, infinity is not a number. Neither is useful for analyzing relationships, and neither can be used as cogs in algebraic proofs.

Fundamentally, however, this is more an issue of applying logic incorrectly in a proof than it is about what happens when you try to divide by zero.

As @Reeii Education pointed out, a similar apparent contradiction appears when multiplying by zero even in a context where the algebra is correct. If [MATH]x = 0[/MATH], then [MATH]1x = 0[/MATH] and [MATH]2x = 0[/MATH], meaning [MATH]1x = 2x[/MATH] and ultimately [MATH]1 = 2[/MATH]. In this situation, all the working-out is correct, but the result is still wrong. Doing the algebra is only part of the proofs process: logic is another important part that is often forgotten or overlooked.

One of my favorite puzzles demonstrates this concept. It goes like this:
A right triangle has a hypotenuse of 8 and an elevation from the hypotenuse of 6 (distance from the hypotenuse to its opposite vertex). What is the area of the triangle?
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The typical approach to solving this employs the simple formula for the area of a triangle: [MATH]A = \frac{1}{2}bh[/MATH], where [MATH]b[/MATH] is the base of the triangle and [MATH]h[/MATH] is its height. Here, the base is 8 and the height is 6, making the area 24.

However, the parameters here are invalid. The maximum possible elevation from the hypotenuse of a right triangle occurs in the 45-45-90 case, where the elevation is half the length of the hypotenuse. In the puzzle, the elevation was said to be 6, which is greater than half of 8, meaning the described triangle is impossible.

The formula is correct, and it is used correctly. Nonetheless, the answer is bogus because the logic is wrong.
__________

On a less related note... Another example of using the wrong logic can occur when using the formula for the infinite geometric series. Each term in the series is found by multiplying the previous term by a common ratio. Where [MATH]a[/MATH] is the initial term and [MATH]r[/MATH] is the common ratio, the sum is found like this:

[MATH]x = \sum_{n=0}^{\infty} ar^{n}[/MATH]​

It's clear that the ratio needs to be such that [MATH]-1 < r < 1[/MATH], or else the sum will trend towards an infinity. But what happens when we produce a formula to solve for this?

[MATH]x = ar^0 + ar^1 + ar^2 + ar^3 + ...[/MATH]
[MATH]rx = ar^1 + ar^2 + ar^3 + ...[/MATH]
[MATH]x - rx = (ar^0 + ar^1 + ar^2 + ar^3 + ...) - (ar^1 + ar^2 + ar^3 + ...)[/MATH]
[MATH]x - rx = a[/MATH]
[MATH]x(1 - r) = a[/MATH]
[MATH]x = \frac{a}{1 - r}[/MATH]​

In this way, we are able to derive a discrete formula to solve for an infinite series. At a glance, it looks like we can use ratios outside of the [MATH](-1, 1)[/MATH] range. For instance, if [MATH]a = 3[/MATH] and [MATH]r = 4[/MATH], we get the following:

[MATH]x = \sum_{n=0}^{\infty} 3(4^n) = 3 + 12 + 48 + 192 + ...[/MATH]
[MATH]x = \frac{3}{1 - 4} = -1[/MATH]
[MATH]\therefore 3 + 12 + 48 + 192 + ... = -1[/MATH]​

The formula is correct, and it is used correctly. Nonetheless, the answer is bogus because the logic is wrong. It's easier to check this time, though: it's clear that we can't add successively larger positive numbers and have them converge to a negative value.

... Wait, doesn't this seem a bit familiar? Isn't this pretty much what happened with the Riemann zeta function?
 
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