Question

For all positive integers \(n\), \show that the decimal expansion of \(\large \dfrac{1}{n}\) terminates if and only if \(\large n = 2^a5^b\)

\(a,b\ge 0\)

Answer 1

hmm so n needs to be >1 right?

Answer 2

yes assume n is positive integer :)

Answer 3

anyways if a abd b>0 then 2^a*5^b>0 too

Answer 4

No, not necessarily, there's no reason why a and b can't both be 0.

Answer 5

oh yeah

Answer 6

hmm is a0 and b=0 then 2^a*5^b=1

Answer 7

if a=0 and b=0 then 2^a*5^b=1

Answer 8

Yeah, so as long as we are calling that 1.0 it is the most rapidly terminating decimal of all of them... (or maybe technically a counter example)

Answer 9

but le me think

Answer 10

there cant be a case where 1/0. thats impossible

Answer 11

Since a,b are greater than or equal to 0, this is the minimum. To get zero in the denominator we would need negative infinity to be in the domain for a and b. \[\Large \frac{1}{2^05^0}=1 \]

Answer 12

n cannot be 0

Answer 13

2^0*5^0=1 thus n>0

Answer 14

right, let me add that to the main q so it becomes more clear :)

Answer 15

then 1/n<=1

Answer 16

\[1/n \le1\]

Answer 17

@ganeshie8 thats hard

Answer 18

so we just got that 1/n<=1 but what this gived us? hmmmmm

Answer 19

\[0<1/n \le1\]

Answer 20

below doesnt terminate
```
1/3 = 0.3333333...
```

below terminates
```
1/2 = 0.5
```

Answer 21

hmmmmm gsh

Answer 22

@dan815

Answer 23

1/2=.5 and 1/5=.2 and are terminating decimals, and I can't see a reason why you would believe multiplying two nonrepeating decimals would create an infinitely repeating decimal. It seems to work because 2 and 5 are the only divisors of 10. I guess this probably doesn't count as a real proof.

Answer 24

Ahh nice that is a legitimate proof in one direction xD
what about the converse ?

Answer 25

Well, 2 and 5 are the only divisors of 10 so any other choice won't work haha, I guess that's also sort of weak.

Answer 26

\(\Rightarrow\)

1/2 = .5 terminates so (1/2)^a terminates

1/5 = .2 terminates so (1/5)^b terminates

so their product 1/(2^a5^b) also terminates

Answer 27

yeah for the converse, if there is some easy intuition of why only the combinations of divosors of 10 will work..

Answer 28

I guess my alternate thing was playing with this, if b>a then \[\Large \frac{1}{2^a5^b}=\frac{2^{b-a}}{10^b}\] and it should terminate since it divides evenly haha I don't know

Answer 29

Haha I don't really know how to prove things, I mean if it's true, it's true and there are many ways you could not believe that this is true haha

Answer 30

that looks more neat
\[\Large \frac{1}{2^a5^b}=\frac{2^{b}5^a }{10^{a+b}} \]

still this wont help in proving converse i think...

Answer 31

converse : \(\dfrac{1}{n}\) terminates \(\implies \) \(n = 2^a5^b\)

this has to address other primes also i guess

Answer 32

Well powers of 2 and 5 are the only numbers that are evenly divided by powers of 10, everything else is relatively prime so nothing else will work. I guess that's really all I am saying is my proof of the converse but I guess that's not quite enough hmmm

Answer 33

you're saying \(n \nmid 10^a\) when \(n\) is made up of other primes ?

Answer 34

that looks good enough to me :)

Answer 35

i believe all other proofs boil down to saying the same thing. .

Answer 36

Haha so is that it? I feel like I kinda avoided stuff, like why am I not expected to prove that two nonrepeating decimals always produces a nonrepeating decimal haha. I don't know how to prove such a thing, but I wouldn't see a reason to. It would be like proving two integers can never multiply together to make an irrational number or something haha.

Answer 37

i think we take them for granted in number theory, we wry about such petty stuff only in analysis :P

Answer 38

Ok good, but yeah even so, I feel like there's sort of an endless pit of possible things we could prove, so I don't know what's exactly acceptable in the world of number theory XD