# Anything divided by 0 is now defined(if I'm right)

Discussion in 'Physics & Math' started by Muhammad Umer, Feb 14, 2017.

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Im Muhammad Umer from Lahore, Pakistan . I'm a student of O'level and gave a thought to X/0.
When we say 1/0.1, ans is 10. when we say 1/0.001, ans is 1000. As the denominator gets smaller, our value of ans increases. So when we say 1/0, logically the ans should be ~(infinity), or more appropriately 1x 10^~ (^ = raise to power). Hence, 2/0=2x10^~,
3/0=3x10^~,
-1/0=-1x10^~,
-2/0=-2x10^~ and so on. But a problem arises when we divide 0 by 0.
First if we all agree that 2/0=2x10^~, multiply both sides by 0, which gives:
0x2/0=0x(2x10^~). Think 2x10^~ as a very big number, when it is multiplied by 0, ans is 0. But on left side 0 denominator of itselfe:
2x(0/0)=0, take two on the other side and get 0/0=0,
or alternatively, assume 0/0=y, 2y=0 and y=0.
Note that 0/0 is only case in which you cannot simply cut numerator by denominator and you can also not take 0 in denominator as such and have to properly multiply both sides by 0.
If I'm wrong then kindly point out my mistake and convince me about it too. But if I'm right then I need some help spread the words. Thanks.

3. ### originTrump is the best argument against a democracy.Valued Senior Member

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This is probably beyond what you have studied, so I think this was a good exercise you have tried.

You are saying that [infinity x 0 = 1] and I do not think anyone can agree with that.

In your example of 1/0.1, it is true that as the denominator approaches 0 the result approaches infinity, but the fraction is undefined when the denominator is 0.

This has to do with the concept of limits. You will learn about these later in your education.

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7. ### rpennerFully WiredStaff Member

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When you say X/Y = Z, you are talking about 3 numbers such that X = Y × Z.

That is not logic. You seem to be claiming that because 0 is the limit of the infinite sequence {1, 0.1, 0.01, 0.001, ... } that 1/0 is the limit of the infinite sequence { 10, 100, 1000, ... } or $10^{\infty}$. This does not follow. For 0 is the limit of many infinity sequences such as {1, 1/2, 1/3, 1/4, 1/5, ... } and { 1, −0.5, 0.25, −0.125, 0.0625, ... } which your claim suggests that means $10^{\infty} = \infty = (-2)^{\infty}$. So I put it to you, you are not using the rules of math and logic alone, but are making a hidden assumption.

That assumption is that ∞ is a number, rather than a concept.

For if ∞ is a number, then we need to know how to do arithmetic with it. What is ∞ + 1 ? Let us figure it out by the method of common denominators, where 1 is any non-zero number divided by itself because this method works for any rational number Y/Z:

Y/Z + 1 = Y/Z + X/X = ( Y × X + Z × X) / ( Z × X ) = (Y + Z)× X / ( Z × X ) = (Y + Z)/ Z ✅

∞ + 1 = 1/0 + 1 = 1/0 + X/X = ( 1 × X + 0 × X)/ ( 0 × X ) = ( X + 0 )/ ( 0 ) = X/0 = X × ∞ ⚠️

This mean ∞ + 1 = −∞ and ∞ + 1 = ∞ and ∞ + 1 = 2 ∞ which seems to contradict your claim that −1/0 and 1/0 and 2/0 are different numbers.

So claiming 1/0 = ∞ is a problem because 1 is a number and 0 is a number, but ∞ is not a number. Because ∞ is not a number, 0 × ∞ is not a number and so 1/0 = ∞ means X/0 = ∞ and since we have that X/0 = ∞ means −1 / 0 = ∞ AND 1 / 0 = ∞ AND 2 / 0 = ∞ then you are saying that −1 = 0 × ∞ AND 1 = 0 × ∞ AND 2 = 0 × ∞ which is to say −1 = 1 = 2 which is clearly not correct.

There is, however a similar concept.
Let ε be a positive number, smaller than any positive number. Then 1 > 1/2 > 1/3 > 1/4 > ... > ε > 0. We call such non-standard numbers, infinitesimals.
Because it's a number, ω = 1/ε is a non-standard number larger than any standard integer.
So we have ... < −ω⁴ < −ω³ < −ω² < −ω < ... < $-10^{50}$ < −10 < −1 < −1/64 < −ε < −ε² < ... < 0 < ... < ε² < ½ε < ε < 2ε < ... < $10^{-50}$ < 1 < 100 < ... < ½ω < ω < 5 ω < ω² < ω³ < ...

These are the hyperreal numbers. https://en.wikipedia.org/wiki/Hyperreal_number

The general form of a hyperreal number is $\sum_{k=a}^{b} c_k \omega^k$. They can be used to solve some problems in calculus in a non-standard way, but aren't needed because the same results are achieved in standard calculus.

But division by zero is still not allowed.

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You are saying that [infinity x 0 = 1] and I do not think anyone can agree with that.

No,no .... actually I said (infinity x 0=0, not one). But I had no idea about limits so thank.

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comment by reppener was a bit complicated but got that too, thanks guys.

10. ### originTrump is the best argument against a democracy.Valued Senior Member

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You wrote: So when we say 1/0, logically the ans should be ~(infinity)
Which means--> infinity x 0 = 1.

No problem. Keep exploring and questioning, that is how you gain knowledge!

11. ### DaveC426913Valued Senior Member

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You guys have covered a lot but I don't think I've seen it explained in the most concise way I've heard.

Division is defined as the inverse of multiplication.

There is no number n that, when multiplied by 0, will give n.

So, literally, 1 divided by zero is ... no number.

12. ### rpennerFully WiredStaff Member

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There is no number n that, when multiplied by 0, will give 1, or any other non-zero number.

13. ### DaveC426913Valued Senior Member

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I used the wording from the Wiki article.

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My teachers still appreciate my efforts, though I was not correct.

15. ### CounterRegistered Senior Member

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(((n+1)/(0+1))-1)=n

10/((10/10)/10)=10/1/10=10/0.1=100/1=100

16. ### DaveC426913Valued Senior Member

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Not quite.

10/1/10=1

What you meant was

10/((10/10)/10)=10/(1/10)=10/0.1=100/1=100