Are We Rounding?

Algebra Level 2

True or False?

$0.49999\ldots = 0.5$

True False

17 solutions

Otto Bretscher
Jan 4, 2016

To avoid the kind of arguments and doubts we see on this page, it helps to remember how the real numbers are actually defined.

One approach, by " Dedekind cuts " (Dedekind Schnitt), defines a real number as a partition of the rationals into two non-empty subsets $A$ and $B$ such that $a<b$ for all $a$ in $A$ and $b$ in $B$ , and such that $A$ has no largest element. Now let $A$ be the set of all rational numbers $a$ such that $a<0.499...999$ for some (finite) number of 9's, or, equivalently, $a<\frac{1}{2}-10^{-n}$ for some positive integer $n$ . We can verify that $A$ and its complement $B$ define a Dedekind cut, that is, they define a real number, which we choose to denote by $0.4999...$ . The equation 0.4999... = 0.5 follows since a rational number is $<\frac{1}{2}$ if and only if it is $<\frac{1}{2}-10^{-n}$ for some $n$ .

Moderator note:

Great approach for explaining how to derive this real number, through the understanding of Dedekind cuts. Unfortunately, as you mentioned, most people are unfamiliar with this terminology, which is why we avoided such an explanation in the misconception wiki page.

I'm new to Dedekind, can you explain this line:

We can verify that $A$ and its complement $B$ define a Dedekind cut, that is, they define a real number, which we choose to denote by $0.4999...$ . The equation 0.4999... = 0.5 follows since a rational number is $<\frac{1}{2}$ if and only if it is $<\frac{1}{2}-10^{-n}$ for some $n$ .

What is the reasoning behind all these? It looks to me that you just form fragments of sentences with no connection whatsoever...

Can you enlighten me on this issue? Thanks...

Pi Han Goh - 5 years, 5 months ago

So, why is A,B a Dedekind cut? There are four things to verify: They form a partition of the reals, they are both non-empty, a<b whenever a is in A and b is in B, and A has no largest element. Do you have doubts on any of those?

Otto Bretscher - 5 years, 5 months ago

Pardon me, but since I'm learning mathematical analysis on my own, I'm not sure why this mechanics actually works.

The issue I'm having here is that yes, I understand that you need these conditions to be verified (no doubt on any of those) to make sure that they are in a Dedekind cut (as you have provided in the wiki link), but I can't seem to understand the mechanics behind it at all.

The way I interpret what you said (and what the wiki said) is that you need 4 seemingly random constraints to be fulfilled so that you can declare it to be a Dedekind cut, hence my question, "What is the reasoning behind all these? It looks to me that you just form fragments of sentences with no connection whatsoever..."

I even open up the book, Principles of Mathematical Analysis by Rudin , and only it only talks about Dedekind on page 21, where they're trying to prove a seemingly irrelevant theorem:

There exist an ordered field $R$ which has the least-upper-bound property. Moreover, $R$ contains $Q$ as subfield.

So I'm really lost here.

Pi Han Goh - 5 years, 5 months ago

@Pi Han Goh – Dedekind describes this process in his booklet "Stetigkeit und irrationale Zahlen,", "Continuity and irrational numbers."

We assume that the rational numbers have been defined, and now we are trying to define the real numbers based on the rationals. Intuitively, we want every real number $x$ to be uniquely determined by the rationals $q<x$ , so we go ahead and define a real number in terms of the rationals that are smaller and those that are larger (or equal)... these two sets define the Dedekind cut.

I think this is an ingenious way to define the reals... do you know a better way?

By the way... did you try this problem ? It's a good problem... I can't take credit for it since @Aareyan Manzoor suggested it. You are among just a handful of people on this site who could write a good solution. (You will know what to do with the cryptic hint, once you decipher the Cyrillic.)

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – .

We assume that the rational numbers have been defined, and now we are trying to define the real numbers based on the rationals. Intuitively, we want every real number $x$ to be uniquely determined by the rationals $q<x$ , so we go ahead and define a real number in terms of the rationals that are smaller and those that are larger (or equal)... these two sets define the Dedekind cut.

How about the irrational numbers? Are you saying that irrational numbers are not real numbers?

Yeah, I couldn't solve that problem, but I haven't given up on it yet.

Pi Han Goh - 5 years, 5 months ago

@Pi Han Goh – The whole point of Dedekind cuts is to define the irrational numbers. To define $\sqrt{2}$ , for example, we let B be all positive rational $x$ with $x^2>2$ and A the compliment.

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – Okay finally understood. I misinterpreted $x = (A,B)$ as a coordinate point. Let me see if I can post a question on Dedekind....

Pi Han Goh - 5 years, 5 months ago

@Otto Bretscher – Do you know a simple proof of Dedekind cuts?

Pi Han Goh - 5 years, 5 months ago

@Pi Han Goh – What you mean by proof? It's a definition...

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – Or: "why does it work?"

Pi Han Goh - 5 years, 5 months ago

@Pi Han Goh – That would be a very informal question: "work" in what sense? You assume that we have some intuitive sense of what the real numbers are.

Dedekind defined the set of real numbers as the set of Dedekind Cuts.

As this very problem shows, most people's perception of the real numbers is really quite fuzzy.

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – Hmmmm.... I must be missing something fundamental here again. Let me put on my thinking cap.

Pi Han Goh - 5 years, 5 months ago

@Otto Bretscher –

I think this is an ingenious way to define the reals... do you know a better way?

Haha, come to think of it, I was just finding an excuse to simplify your explanation. Turns out, it really is the simplest.

Pi Han Goh - 5 years, 5 months ago

@Pi Han Goh – Many school books "define" the reals as non-terminating decimals... but this very problem shows that this definition does not quite work.

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – Relevant video . Warning: Slight coarse language.

Thanks for the conversation!

Pi Han Goh - 5 years, 5 months ago

@Pi Han Goh – Great video indeed! Thanks!

My "motto" is relevant too: Das Wesen der Mathematik liegt in ihrer Freiheit. It's all what we say it is.

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – i have deciphered it... should be matter of time before i solve it... i didnt notice that problem before for some reason....

Aareyan Manzoor - 5 years, 5 months ago

@Aareyan Manzoor – Great! I'm looking forward to another one of your clear and elegant solution...

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – i got stuck despite the hint... i will try this again later.

been trying this for 50 mins!

Aareyan Manzoor - 5 years, 5 months ago

@Aareyan Manzoor – You have only yourself to blame... you suggested the problem ;)

I will make the hint a little stronger!

Otto Bretscher - 5 years, 5 months ago

New question : How do I prove that $\sqrt 2$ is irrational by Dedekind's cut? How about $\ln3 , e , \pi$ , etc? Can we prove/disprove the irrationality of all numbers by Dedekind's cut?

Pi Han Goh - 5 years, 5 months ago

Does this only work for a finite value for $n$ ?

Joshua Nesseth - 4 years, 5 months ago

if you multiply 0.999... by 10 wouldn't it take a zero off the end of x and be infinity -1 repeating digits? because if that were the case 10x -x <9 which then divided by nine is <1

Cole Maney - 3 years, 2 months ago

Rishik Jain
Jan 5, 2016

Let $x = 0.49999.....$

$10x = 4.9999......$

$10x - x = 4.9999.... - 0.49999.....$

$9x = 4.5$

$x = \dfrac{4.5}{9}$

$x = \dfrac{1}{2}$

$x = 0.5$

$\large \boxed{0.49999.... = 0.5}$

We are not going for memorised thing!

Dhruv Joshi - 3 years, 4 months ago

Karthick Shiva
Jan 2, 2016

0.49999999..... = 45/90 = 1/2 = 0.5.

how is it equal to 45/90 ??

Chirag Gupta - 5 years, 5 months ago

$x = 0.499 \ldots$

$10x = 4.99 \ldots$

$10x-x = 4.99 \ldots - 0.499 \ldots$

$9x = 4.5$

$90x = 45$

$x = \frac {45}{90}$

Manuel Kahayon - 5 years, 5 months ago

assume x=0.04999999… then 10x=0.4999999…. 9x=0.4499999999… 90x=4.499999999…. x=0.0499999999..

which is what we assumed x to be in the first place, thus it is more accurate ( no rounding ) ( plus the equal sign in the question shows that it is " exactly" equal ... which is definitely not true)

point is : what you did in your proof, is rounding to one decimal ( by assuming x=0.49999... and multiplying by 10 in your steps) my point here, it could go both way, depending on you assumption of "x" at the beginning , hence it is not a "true" or "false" answer, it merely depend on: how many decimal points you are willing to round to ...

Ziad Osman - 5 years, 5 months ago

The logic of this proof (and other corollary ones) is questionable.

You purport to multiply a given number by ten; let us use pi, instead. If you were to multiply pi by ten, you would simply be left with $10 \pi$ . This is because $\pi$ is an irrational number (i.e. it cannot be transformed into a ratio form).

Implicit in you response is that any variant of $x.x \dots x \overline{9}$ would is, specifically speaking, a rational number [i.e. $x.x \dots x \overline{9} \, \in \, \mathbb{Q}$ ].

This is, I would argue, a reversal of the implication of rational numbers--specifically, $\frac{1}{3} \, \mapsto \, 0.\overline{3}$ . It is easy enough to prove this ratio can be "represented" by a repeating decimal (via mathematical induction); the reverse, however, is not (necessarily) implied.

In other words, you first have to PROVE $10(0.\overline{9}) \, = \, 9.\overline{9}$ , by FIRST proving either $[ 0.\overline{9} \, \in \, \mathbb{Q} ]$ OR $[10^x (0.\overline{9} \, = \, \exists y \: \text{for} \, \forall \, x ]$ .

I would argue the ability to multiply repeating decimals derives from their corollaries in rational numbers (i.e. fractional avatars), RATHER than their numerical existence--which, it should be remembered, is NOT equivalent to the finite numbers we are used to dealing with.

Unless I missed something?

Joshua Nesseth - 4 years, 5 months ago

@Joshua Nesseth – We have to prove that $10(0. \overline{9}) = 9. \overline{9}$ ?

$\large 0. \overline{9} = \displaystyle \sum_{i=1}^\infty \frac{9}{10^i} \implies 10(0. \overline{9}) = 10(\displaystyle \sum_{i=1}^\infty \frac{9}{10^i})$

$\implies 10(0. \overline{9}) = \displaystyle \sum_{i=1}^\infty 10\frac{9}{10^i} = \displaystyle \sum_{i=1}^\infty \frac{9}{10^{i-1}} = \displaystyle \sum_{i=0}^\infty \frac{9}{10^i} = 9. \overline{9}$

Done.

Manuel Kahayon - 4 years, 5 months ago

@Manuel Kahayon – [In case you don't want to read the whole thing--it's the initial equivalence I have a problem with; it begs the question.]

You are assuming (as I had intended to preempt in the post) from the outset that

$0.\overline{9} \, = \, \displaystyle\sum_{i=1}^{\infty} \frac{9}{10^i}$

--an assertion that, while I do not deny COULD be possible, is not proven.

The former, as it has been pointed out may times, is a number; the latter, an infinite sum. While it is true that convergent infinite sums can be thought of as numbers, it can be shown (again, as it has many times in this thread) that

$\displaystyle\sum_{i=1}^{\infty} \frac{9}{10^i} \: \text{converges to} \: 1$

It still remains unproven that:

$\displaystyle\sum_{i=1}^{\infty} \frac{9}{10^i} \: \text{converges to} \: 0.\overline{9}$ .

You cannot use a sum that converges to a given number (here, $1$ ) to prove that another number (here, $0.\overline{9}$ ) is equal to it.

I don't deny the "idea" of a sum of infinitely diminutive products of $9$ isn't intuitively appealing, but it cannot be ASSERTED without proof of it's initial equality to the desired number--a proof that is, here, assumed rather than asserted [i.e. it begs the question].

Not to mention, isn't this entire exercise meant to re-orient people's assumptions of equivalence? So, then, why is it acceptable to use colloquial equivalence when asserting a sum's equivalence, but not when asserting a non-equivalence of two numbers?

[Beyond all this, I would argue a convergent infinite sum is--ontologically--quite different from an irrational number; but that, unfortunately, is a discussion for a different day...]

Joshua Nesseth - 4 years, 5 months ago

@Joshua Nesseth – $\displaystyle \sum_{i=1}^\infty \frac{9}{10^i} = 0.9+0.09+0.009+0.0009+... = 0.999\cdots = 0.\overline{9}$ .

Manuel Kahayon - 4 years, 5 months ago

@Manuel Kahayon – I understand the "inuition" behind the statement...as I have pointed out above.

Also, as I have pointed out above, it can be shown that

$\displaystyle\sum_{i=1}^{\infty} \frac{9}{10^i} \, = \, 1$ ;

this is not (necessarily, at least) the same as $0.\overline{9}$ .

I am not saying, of course:

$\displaystyle\sum_{i=1}^{\infty} \frac{9}{10^i} \, \neq \, 0.\overline{9}$ ;

that is, I am not directly precluding the possibility.

I posit only that the equivalence cannot be simply assumed. As I have laid out before, I would argue $[ 0.\overline{9} \: \in \: \mathbb{Q} ]$ , in large part due to the fact one cannot assign a corollary ratio value to $0.\overline{9}$ without first assuming $[ 0.\overline{9} \: = \: 1]$ (&, therefore $[ 0.\overline{9} \: = \: \displaystyle\sum_{i=1}^{\infty} \frac{9}{10^i} ]$ ).

Again, I do not speak to it's falsity, but merely its lack of proof.

Joshua Nesseth - 4 years, 5 months ago

the nearest integer to 0.5 is also taken to be 0. so, the ans must be 0.

Aryan Chugh - 5 years, 5 months ago

@Aryan Chugh – By convention 0.5 rounds to 1 not to 0. This is just a rule to learn.

Karthick Shiva - 5 years, 5 months ago

@Karthick Shiva – according to the nearest integer function (nint), nearest integer to 0.5 is 0, you can google the graph if you want to.

Aryan Chugh - 5 years, 5 months ago

@Aryan Chugh – I once learned in physics: Any .5 number which preceded by an even number should be rounded down. And any .5 number which preceded by an odd number should be rounded up. Since in this case 0 is even, then I thought I should round it down to 0.

Henny Lim - 5 years, 5 months ago

@Henny Lim – Yes, exactly, that's what I learned too... this is the standard widely used today by mathematicians and in most mathematical software. But as you can see on this page, not everybody agrees ;)

Otto Bretscher - 5 years, 5 months ago

@Aryan Chugh – Uhh, well, it can actually be 0 or 1, but 1 is more accepted by society. One reason for rounding up at 0.5 is that for positive decimals, only the first figure after the decimal point needs be examined. For example, when looking at 17.5000…, the "5" alone determines that the number should be rounded up, to 18 in this case.

Manuel Kahayon - 5 years, 5 months ago

@Manuel Kahayon – first you rounded off 0.49999... to 0.5 then you say 0.5 is close 1 . but 0.4999999.... is definitely closer to 0 than to 1.

Aryan Chugh - 5 years, 5 months ago

@Aryan Chugh – Umm, we didn't round off 0.49999... to 0.5, 0.4999... is actually equal to 0.5... I dont know if you'll actually believe it, but it's true... well, imagine 0.5 minus 0.1, 0.01, 0.001 and so on..

$0.5-0.1 = 0.4$

$0.5-0.01 = 0.49$

$0.5-0.001 = 0.499$

$0.5-0.0001 = 0.4999$

This is somewhat like limits in calculus, but as you can see as 0.5 gets subtracted by 0.1 raised to some number x, as x gets closer and closer to infinity, then 0.4999... gets more and more 9's... Since 0.4999... has infinite nines, then we can say that $0.4999 \ldots = 0.5 - 0.1^{\infty}$

Since $0.1^{\infty}=0$ , then $0.4999 \ldots = 0.5$

Manuel Kahayon - 5 years, 5 months ago

@Manuel Kahayon – I don't think you can say stuff like 0.1^\infty =0. You can say lim(0.1^x)x->\infty=0

Denis Denis - 5 years, 5 months ago

@Manuel Kahayon – Since 4.999.... > 0.5 and since any number > 0.5 must be rounded to 0 Then 0.4999.... Must rounded to 0 and Can't be rounded to 1

Ramy Mohsen - 5 years, 5 months ago

@Aryan Chugh – Hey friend , $0.4999...$ is exactly equal to $0.5$ and it is not rounded to $0.5$ . Of course , $0.5$ is close to $0$ as it is to $1$ . But , by convention , or due to some reason , mathematical society has decided that whenever you have to round off numbers like $4.5$ , it must be rounded to $5$ and not $4$ .

I hope you get the point now :)

Nihar Mahajan - 5 years, 5 months ago

@Nihar Mahajan – No, mathematical society has not decided that, just some old text books and out-of-touch schools. The IEEE standard today is to round to the closest even number. Try round(0.5) on WolframAlpha!

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – The rounding IEEE rule that you are referring to is just one of five rounding rules the IEEE has published. The IEEE publishes rules for computational handling of floating point arithmetic. It is not the authoritative body on Mathematical conventions. Your comment about Out-of-touch schools is inaccurate, misleading (especially for students) and demeaning to Mathematics educators in schools worldwide.

Andrew Bone - 5 years, 5 months ago

@Andrew Bone – I count two IEEE rules for "rounding to the nearest" (not five), namely, "rounding to even" and "rounding away from zero", with rounding to even being the "recommended default".

I'm not saying that "rounding to evens" is the only way to do it, although it appears to reduce the accumulation of round-off errors. I think any math educator should make it clear to her students that in some cases there are no fully agreed-upon conventions as to how things are done; this is certainly one of those cases. The teacher may say (as I do in my classes) that in his class things are done in such-and-such way, but others do it differently. When it comes to rounding half-integers, a student may notice that WolframAlpha rounds to evens while Python rounds up.

I have no tolerance for math educators who teach some arbitrary rules to their students and present them as "holy writ"... the effects of such teaching can be seen on this page: People (including the two of us) arguing about conventions when they could be doing real math. ("This is just a rule to learn"..."4.5 must be rounded to 5 and not 4")

Otto Bretscher - 5 years, 5 months ago

@Nihar Mahajan – Oh, my, do you actually have the time to read through all of the comments on all of the problems on brilliant? I mean, you're pretty much everywhere on brilliant XD... just curious XD

Manuel Kahayon - 5 years, 5 months ago

@Manuel Kahayon – That may be coincidence. I am always there to help people. Its my duty!

Nihar Mahajan - 5 years, 5 months ago

@Nihar Mahajan – Oh... So dedicated... That makes me curious about you... maybe we'll meet again sometime on another problem... :)

Manuel Kahayon - 5 years, 5 months ago

@Aryan Chugh – No, 0.499... does not ROUND to 0.5; 0.499... is EQUAL to 0.5! One of the commentators above (Manuel Kahayon) has given an elegant mathematical proof that it is so. The question remains which integer 0.5 (or 0.499...) is closer to, and it is very easy to prove that they are equally distant. Zero point five plus or minus 0.5 is 0 or 1.

Peter O'Donnell - 5 years, 5 months ago

@Manuel Kahayon – 1 isn't "more accepted by society", just by some out-of-touch math teachers ;)

Otto Bretscher - 5 years, 5 months ago

That technic is not accurate, because if you times ten there is supposed to be 'one less 9' at the end of the number. like 0.499999999999999999999999999999990

Cisco Wicaksana - 5 years, 5 months ago

@Cisco Wicaksana – But, there is an infinite number of 9's at the end of the number, and since infinity minus one is still infinity, it works... Actually, you can convert all repeating decimals to fractions using this technique...

Manuel Kahayon - 5 years, 5 months ago

@Manuel Kahayon – Before you can settle this issue, you have to ask yourself: That is the definition of a real number in the first place? People tend to get into these kinds of arguments when they are not using solid definitions.

Otto Bretscher - 5 years, 5 months ago

False assumptions lead people to think 0.499... = 45/90

Toni Almeida - 2 years, 11 months ago

Thanks. To remove ambiguities, we've completely revamped the problem. Those who previously answered this problem will marked as correct.

Brilliant Mathematics Staff - 5 years, 5 months ago

Thank you, i wish i saw this before i deleted my posts. I could have sworn the original post was changed.

j c - 5 years, 5 months ago

The solution is still wrong. Rounding has nothing to do with the problem. Sure, the title asks if you know how to round but the actual problem just makes a statement and asks if it is true or false. Assuming the ... means the 9 repeats to infinity then the statement is true.

Thomas Jones - 5 years, 5 months ago

Omg, never mind. Who puts False as the first choice of a true/false question? I guess I should read the answers better. Lol

The problem still has nothing to do with rounding. 0.49999... does not equal 0.5 because of rounding. It is, by definition, EXACTLY 0.5.

Thomas Jones - 5 years, 5 months ago

i dont think that this solution is correct ......

Aryan Chugh - 5 years, 5 months ago

misleading

Sanjoy Roy - 5 years, 5 months ago

the jump from 4.999.... to 45/90 is an assumption that 4.999.. =5

Cole Maney - 3 years, 2 months ago

0.49999999=(49-4)/90=45/90=1/2=0.5 0.49999999= x=0.499999..... 10x=4.9999999.... 9x=4.5 x=0.5

. . - 10 months, 4 weeks ago

0.499999... comes to be approximately 44.1/90 which is not equal to 0.5. Hence it's false and not true

Karthik Sridhar - 5 years, 5 months ago

why is the solution marked as false? I answered true and got it wrong...

Davin Shaun - 5 years, 5 months ago

We have checked with the system and already have updated all previous attempts as correct.

Brilliant Mathematics Staff - 5 years, 5 months ago

@Brilliant Mathematics – thanks! btw how do you guys respond so fast? Do you just filter all comments with 'false' in them or? I'm guessing you have some sort of code to filter the comments most likely to be displaying wrong answer complaints or something. Am i right? or do you guys just all read this stuff a lot?

Davin Shaun - 5 years, 5 months ago

Mohneesh Khaneja
Feb 8, 2016

Just like 1/3 = 0.33333... ==> Multiply both sides by 3 to get 1 = 0.99999... By similar argument 0.499999... = 0.4 + 0.099999... = 0.4 + 3 (0.033333...) = 0.4 + 3 (1/30) = 0.4 + 0.1 = 0.5

Jonah Burian
Jan 10, 2016

.333...=1/3

.666...=2/3

.999...=3/3

So: .999...=1

4.999...= 4+.999... = 4+1 = 5

Devanshi Shah
Jan 4, 2016

I still dont understand. 4.9999999999999 is almost 5.0 but not exactly, right? I already read the article it looked like baloney to me.

It is what is called ‘approaching’.

Bao NGUYEN - 5 months, 3 weeks ago

In this question , a similar concept is explored and there are arguments of $0.\overline{9}$ , both "close to" and "exactly"

Kay Xspre - 5 years, 5 months ago

Infinity Mathematics
Apr 3, 2018

Another approach to this problem:

We can express the number $0.4999\dots$ as the geometric sequence $\frac{9}{10^2} + \frac{9}{10^3} + \frac{9}{10^4} + \frac{9}{10^5} \dots$ plus the value $0.4 = \frac{2}{5}$ . Therefore, we can express $0.4999\dots$ as $0.4 +\frac{9}{10^2} + \frac{9}{10^3} + \frac{9}{10^4} + \frac{9}{10^5} \dots$ . Using the geometric series formula, we have:

$0.4999\dots = 0.4 +\frac{9}{10^2} + \frac{9}{10^3} + \frac{9}{10^4} + \frac{9}{10^5} \dots = 0.4 + \frac{\frac{9}{10^2}}{1-\frac{1}{10}} = 0.4+\frac{\frac{9}{10^2}}{\frac{9}{10}} = 0.4 + \frac{9}{100} \cdot \frac{10}{9} = 0.4 + \frac{1}{10} = 0.4+0.1 = \boxed{0.5}$

So, $0.4999\dots = 0.5$ is true .

Gini Mas
Jan 9, 2016

If 0.999...=1

Then the identity stated in the question should also be true as 0.5 was subtracted from both sides.

Bao Nguyen
Dec 23, 2020

Assuming $\infty$ interacts with itself normally but not with real numbers, 0.9 = $\frac{9}{10}$ 0.99 = $\frac{99}{100}$ 0.999... = $\frac{\infty-1}{\infty}$
If infinity isn’t formalised, take an isosceles triangle. If you fix the 2 identical points but expand the two identical angles to 89.999...°, the third point (C) will be an infinite distance away, which means the angles of A and B are 90°.

. .
Jul 21, 2020

0.5 and 0.49999999999999 are same. Because 0.5-0.499999999999999999=0.00000000000000000000000. They aren't the same, but we say they're the same. That's because above calculation 0.5-0.499999999999999=0.00000000000. In the above calculation, the final of the decimal place is 1 but the others are 0. We can appear that decimal to the repeating decimal. Like 0.0000000000000000000000000000000000000000000000000000000000000000000000000000.................................................................................. final place is 1.

Sujunjie Sun
Mar 16, 2020

0.49999...=0.4+0.1*0.999.... because 0.999....=1, 0.499999...=0.4+0.1=0.5

Code Awesome
Oct 16, 2019

when 4.99999... is rounded it equals 5

Oon Han
Jul 10, 2019

Let $x = 0.4999...$ .

Then, $10x = 4.999...$ .

Also, $100x = 49.999...$ .

Hence, $90x = 45$ .

Finally, $x = 0.5$ .

Thus, $0.4999... = 0.5$ .

Therefore, the answer is True .

Inunnguaq Josephine Morén
Nov 25, 2018

I have always, wonder why juicewrld999 was 999, i get it now. 999=1 <3 :)

Mike Kelsey
Feb 10, 2017

x = .499999...

10x = 4.99999...

100x = 49.99999...

Subtract 10x from left side and 4.99999... from right side, since they are equal.

90x = 45

x = 45/90

x = .5

Ahmad Saad
Nov 16, 2016

0.49999... = 0.4 + 9/100 + 9/1000 + 9/10000 + ...

              = 0.4 + (9/100)(1 + 1/10 + 1/100 + ...)

              = 0.4 + (0.09)(1/{1-0.1} = 0.4 + (0.09)(10/9) = 0.4 + 0.1 = 0.5

Yash Patil
Jan 21, 2016

Definition of real numbers is that there exists infinite numbers between 2 real numbers. Since there exists no number between 0.499999.... and 5 we can say that both the numbers are the same

Are We Rounding?

17 solutions

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