Divide by zero?

[#1procrastinator]

In Apostol's analytic number theory book, one of the properties listed for divisibility is:

n | 0 (every integer divides zero)

and also, something like 'zero divides only zero'

Could someone please explain what the crap he means? I thought you couldn't divide by zero!

[dc302]

In general, dividing by zero gives you infinity, but infinity is generally not included in real/complex number systems so that is why you can't divide by zero.

Also, zero divided by zero is in fact indeterminate--that means you can't really assign it any value, unlike n/0.
Why? Think about the following three ideas:

n/0 goes to infinity

0/n = 0

n/n = 1

What happens if I substitute 0 into the above three expressions? I get three answers, infinity, 0 and 1.

[Truck]

In general, dividing by zero gives you infinity, but infinity is generally not included in real/complex number systems so that is why you can't divide by zero.

Also, zero divided by zero is in fact indeterminate--that means you can't really assign it any value, unlike n/0.
Why? Think about the following three ideas:

n/0 goes to infinity

0/n = 0

n/n = 1

What happens if I substitute 0 into the above three expressions? I get three answers, infinity, 0 and 1.

0/0 = 1.

LOL weird... but I sorta see it .

[abeybaby]

Its because if you take the limit as a->0 of 1/a, then you get +/-infinity, depending on whether you approach from above or below. so 1/0 is underfined.

if you take the limit as a->0 of 1/a^2, you only get +infinity. so 1/0^2 is infinity, but 1/0 is undefined. pretty fun to play with, do it on your CAS!

[dc302]

Its because if you take the limit as a->0 of 1/a, then you get +/-infinity, depending on whether you approach from above or below. so 1/0 is underfined.

if you take the limit as a->0 of 1/a^2, you only get +infinity. so 1/0^2 is infinity, but 1/0 is undefined. pretty fun to play with, do it on your CAS!

And if you work in the plane, or complex numbers, you get a whole 360 degrees of infinity! Or if you work in the 3 dimensions, welllll... But in high maths, it is often regarded that {infinity} is simply any point at infinity, regardless of whether its neg, or pos, or N30W, or whatever.

[Thu Thu Train]

0!

[kamil9876]

In Apostol's analytic number theory book, one of the properties listed for divisibility is:

n | 0 (every integer divides zero)

and also, something like 'zero divides only zero'

Could someone please explain what the crap he means? I thought you couldn't divide by zero!

The thing is that he is looking at "divisibility", the definition of divisibility is that a|b means that there exists an integer c such that ac=b.

So 0|0 means there exists a c such that 0*c=0, does there exist such an integer? of course, in fact any will do. Note we can define "divisibility" without actually doing any division so no we are not dividing by 0.

[dc302]

In Apostol's analytic number theory book, one of the properties listed for divisibility is:

n | 0 (every integer divides zero)

and also, something like 'zero divides only zero'

Could someone please explain what the crap he means? I thought you couldn't divide by zero!

The thing is that he is looking at "divisibility", the definition of divisibility is that a|b means that there exists an integer c such that ac=b.

So 0|0 means there exists a c such that 0*c=0, does there exist such an integer? of course, in fact any will do. Note we can define "divisibility" without actually doing any division so no we are not dividing by 0.

Thanks kamil, I just realised I actually misread the original question lol.

n | 0 (n divides 0) is NOT n/0. It is the other way around (0/n)

n/0 is n divided by 0, 0/n is n dividing 0. The two are opposite. So n | 0  is the same as saying 0/n = an integer.

For example, 1 | 5 because 5/1 = 5 an integer. 2 | 6 because 6/2=3 an integer.

Another example, you may come across a property:

if a | b, and b | c, then a | c. This means if a divides b, and b divides c, then a divides c. Or, b is able to be divided by a, and c is able to be divided by b. Then c is able to be divided by a.

proof: if a | b, then b = an for some n (so b/a = an/a = n integer) and b | c means c = bm for some m for the same reason.

So c = bm = (an)m = a(nm), so c/a = a(nm)/a = nm integer. Which means a | c.

[#1procrastinator]

^ thanks a lot guys, think I understand it now