Irrational numbers - incomprehnsible?

[WW_III_ANGRY]

Hi, I am having an issue with irrational numbers and the term irrational.

Main Entry: 1ir·ra·tio·nal
Pronunciation: \i-?ra-sh(?-)n?l, ?i(r)-\
Function: adjective
Etymology: Middle English, from Latin irrationalis, from in- + rationalis rational
Date: 14th century

: not rational: as a (1) : not endowed with reason or understanding (2) : lacking usual or normal mental clarity or coherence b : not governed by or according to reason <irrational fears> c Greek & Latin prosody (1) of a syllable : having a quantity other than that required by the meter (2) of a foot : containing such a syllable d (1) : being an irrational number <an irrational root of an equation> (2) : having a numerical value that is an irrational number <a length that is irrational>





Are irrational deductions, such as the square root of 2 completely irrational in the sense that it is lacking normal coherence and are incomprehensible?



For example, no matter how many decimals we expand the square root of 2, when we multiply itself it is always 1.9999 ... (with some random numbers at the end depending on the exact expansion)

So how could the square root of 2 be comprehended, how can it exist logically and conceptually in the mind except on a superficial level that it is the number that when multiplies by itself equals two?

Are not irrational numbers irrational in the fullest sense of the word? There is no number, or integer that when multiplied by itself that will equal two, stating an infinite expansion will not suffice because that seems illogical, yes? An infinite expansion is not possible .

Thank you,

 

[Gallant92]

The sqrt of 2 times by itself is indeed 2. Irritational in maths terms just means can not be expressed in a form such as a/b. Where a and b are integers. i.e. an ending or repeating decimal. Just because it can't accurately be expressed as a decimal doesn't make it any less of a real number than a rational number. Just that we(the vast majority) think in decimals.

Hope it has helped somewhat.

 

[WW_III_ANGRY]

The sqrt of 2 times by itself is indeed 2. Irritational in maths terms just means can not be expressed in a form such as a/10^n. Where a and n are integers. i.e. a decimal. Just because it can't accurately be expressed as a decimal doesn't make it any less of a real number than a rational number. Just that we(the vast majority) think in decimals.

Hope it has helped somewhat.

How is it possible to think any other way though? (√2)(√2) = 2 will only blind us to the true dilemma that a number that can be multiplied by itself will not be able to equal 2? Its uncountable to me. Why do math rules apply it as something countable? This is my dilemma. I think it is unreasonable, aside from its practical applications.

 

[gmax137]

If you drew a perfect square, with sides of length exactly 1, the diagonal would be of length exactly sqrt(2).

So, there's a line length that you can comprehend.

 

[WW_III_ANGRY]

If you drew a perfect square, with sides of length exactly 1, the diagonal would be of length exactly sqrt(2).

So, there's a line length that you can comprehend.

I still can't comprehend that because it still seems like an approximation (although as close as we can get without knowing any better).

Am I stupid?

 

[Diffy]

You are using an definition of language and applying it to a mathematical idea. You should be using a definition of mathematics and applying it to a mathematical idea (see how that works?).

You can find a mathematical definition here: http://mathworld.wolfram.com/IrrationalNumber.html

 

[Hurkyl]

Are not irrational numbers irrational in the fullest sense of the word?

No; they are irrational only in the technical mathematical sense.

 

[gmax137]

I still can't comprehend that because it still seems like an approximation (although as close as we can get without knowing any better).

Am I stupid?

What's approximate? If you have a right triangle with equal length sides, the length of the hypotenuse is sqrt(2) times the length of those sides. Exactly. We can even write down the value: $$\sqrt{2}$$

I guess I just don't see what the problem is.

 

[WW_III_ANGRY]

What's approximate? If you have a right triangle with equal length sides, the length of the hypotenuse is sqrt(2) times the length of those sides. Exactly. We can even write down the value: $$\sqrt{2}$$

I guess I just don't see what the problem is.

square root of 2 is a number that when multiplied by itself = 2, although I cannot find that number precisely base 10 somehow it exists in conceptual geometry and is always an approximation based on what we are measuring. If a triangle or pyramid were to exist in physical gauageble reality, we would always have to magnify the end of the hypotenuse towards an infinite magnification that will in turn give us another decimal point due to the irrational value of sqrt of 2, for ever and ever, (amen.)

 

[Angry Citizen]

"Irrational" refers to its inability to be expressed as a "ratio", I.E. fraction. Furthermore, mathematics is strewn with terms that, to an ordinary human being, would be incomprehensible or misconstrued, such as the infamous "imaginary" number with very real applications and uses. Why, if we were to make mathematics adhere to ordinary language, many things would have to change (the word sine, for instance).

 

[mikeph]

The question is not really mathematical but physical. It's true that if you measured the diagonal of a square in real life you have (1) an uncertainty, and even if you didn't you (2) wouldn't have enough paper or ink to write down the result of your measurement.

Re (1): the uncertainty is a physical thing and mathematics isn't concerned in this context
Re (2): even rational numbers have infinite representations (eg. 2 = 2.0 = 2.0000...), we just shorten them for practicality. Just as we write 3.131313... = 3.13 with dots above the last two digits, we write the square root of 2 as sqrt(2). We were always taught it is fine to leave surds in our answers, because you can't get a more accurate representation of that number on a finite piece of paper.

If you think a number like sqrt(2) (diagonal of a unit square) is troublesome try pi (circumference of a unit diameter circle). Both are irrational but sqrt(2) is at least algebraic, meaning it is the solution of a polynomial with rational coefficients (x^2 - 2 = 0). pi is transcendental (cannot be the solution of any such equation), and there are infinitely more transcendental numbers than algebraic ones.I hope I got all that summary correct but it kind of blows my mind. It does seem impossible to actually imagine irrational numbers, and the number of them as well!

 

[some_dude]

I think your questions are good. There is a book called "Measure and the Integral" by Henri Lebesgue (don't worry, it's not high level, doesn't even assume you know what "multiplication" or a "number" are, very readable) that I highly recommend you get. It's available on amazon used for $10 or $20 and deeply gets at the heart of what's puzzling you. (Again, don't let the author's name intimidate you, if you've heard of it, it requires next to no background in math for the 1st hundred or so pages).

 

[WW_III_ANGRY]

"Irrational" refers to its inability to be expressed as a "ratio", I.E. fraction. Furthermore, mathematics is strewn with terms that, to an ordinary human being, would be incomprehensible or misconstrued, such as the infamous "imaginary" number with very real applications and uses. Why, if we were to make mathematics adhere to ordinary language, many things would have to change (the word sine, for instance).

Yes so wouldn't it be irrational to consider the square root of two to actually exist? I don't see how a number that when multiplied itself actually equals 2! It doesn't make sense to me.

 

[CRGreathouse]

Yes so wouldn't it be irrational to consider the square root of two to actually exist?

No. It is perfectly rational to consider the square root of two to exist.

I don't see how a number that when multiplied itself actually equals 2! It doesn't make sense to me.

Why does it not make sense? What tells you that there is no number like that? And what about numbers like 555444804887496890116; do their square roots exist?

 

[WW_III_ANGRY]

No. It is perfectly rational to consider the square root of two to exist.
Why does it not make sense? What tells you that there is no number like that? And what about numbers like 555444804887496890116; do their square roots exist?

It does not make sense to me because no matter how many decimals the square root of 2 is expanded to, it will always be 1.999 with some varying string of numbers at the tail dependent upon how many decimals expanded to, through infinity. Which is why at best it is 1.99...to infinity, not 2.

I don't know about that large number, but the same applies to finding the square root of any number that results in an irrational number, which as I understand is not precise and infinite.

As I understand it also, finding the numerical value in decimals is the most precise method to evaluate the square root of 2. If there is another more precise method can you please let me know? Thanks,

 

[JSuarez]

May I add that perhaps your troubles stem from identifying a number, is this case $\sqrt{2}$, with its representation relative to a base? To "understand" $\sqrt{2}$ you don't need to refer to its decimal expansion.

 

[Condor77]

@WW III ANGRY: does this help: 2^(1/2) times 2^(1/2)= 2?

 

[sutupidmath]

If p is any positive real number, there is a positive real number x such that x^2=p.

The existence of such numbers is proved easily by considering the set

A={z:z is a positive real number and z^2=<p} Indeed, such a number will be x=supA, such that x^2=p.

Replace, p=2, and you see, that there is actually some real number x, such that x*x=2.

I don't see what is so irrational about this.

EDIT: An even more generalized version:

If p is any positive real number and n is a positive integer, there is a unique positive real number x such that x^n=p.

 

[Mark44]

Yes so wouldn't it be irrational to consider the square root of two to actually exist?

OK, I'll bite. The square root of 2 doesn't actually exist in any physical or measurable sense. The example was given earlier of a square 1 unit on each side. It's not possible to make such a square, since we would not be able to measure exactly 1 unit, or even get the angles to be exactly 90 degrees.

But so what? At its heart, mathematics is not about geometric figures that we physically create. It's about the perfect, idealized versions of these objects--points that have zero width and height, lines with zero width. If we could create a perfect square that was 1 foot by 1 foot, the length of each diagonal would be exactly sqrt(2) ft. If you want that as a decimal number, you'll have to settle for a finite length representation. One of the attributes of an irrational number is that its decimal representation is infinitely long, and doesn't have any finite-length repeating patterns.

I don't see how a number that when multiplied itself actually equals 2! It doesn't make sense to me.

It doesn't make sense to you because your understanding of numbers is not very developed. That's something you can fix.

 

[gmax137]

It's not possible to make such a square, since we would not be able to measure exactly 1 unit...

Yes, but that just begs the question, how is 1 any different from $$\sqrt{2}$$ ? You can't even draw a line of length '1'.

edit: how do I make my $$\sqrt{2}$$ show smaller?

 

[WW_III_ANGRY]

OK, I'll bite. The square root of 2 doesn't actually exist in any physical or measurable sense. The example was given earlier of a square 1 unit on each side. It's not possible to make such a square, since we would not be able to measure exactly 1 unit, or even get the angles to be exactly 90 degrees.

But so what? At its heart, mathematics is not about geometric figures that we physically create. It's about the perfect, idealized versions of these objects--points that have zero width and height, lines with zero width. If we could create a perfect square that was 1 foot by 1 foot, the length of each diagonal would be exactly sqrt(2) ft. If you want that as a decimal number, you'll have to settle for a finite length representation. One of the attributes of an irrational number is that its decimal representation is infinitely long, and doesn't have any finite-length repeating patterns.

It doesn't make sense to you because your understanding of numbers is not very developed. That's something you can fix.

The non existence in the physical world is not necessarily my issue.

As I see it, the number 2 does not have a possible value that when multiplied by itself equals two.

Value being defined as:
4 : a numerical quantity that is assigned or is determined by calculation or measurement <let x take on positive values> <a value for the age of the earth>

Which means finding the square root of 2 is Not applicable. ?

 

[WW_III_ANGRY]

Another question I have is, do all numbers have a value?

 

[HallsofIvy]

It does not make sense to me because no matter how many decimals the square root of 2 is expanded to, it will always be 1.999 with some varying string of numbers at the tail dependent upon how many decimals expanded to, through infinity. Which is why at best it is 1.99...to infinity, not 2.

I don't know about that large number, but the same applies to finding the square root of any number that results in an irrational number, which as I understand is not precise and infinite.

As I understand it also, finding the numerical value in decimals is the most precise method to evaluate the square root of 2. If there is another more precise method can you please let me know? Thanks,

Then your difficulty appears to be thinking that number must be "decimals". That is a very limited way of thinking- there is an enormous difference between "numbers" and "numerals" (the symbols we use to represent numbers).

 

[WW_III_ANGRY]

Then your difficulty appears to be thinking that number must be "decimals". That is a very limited way of thinking- there is an enormous difference between "numbers" and "numerals" (the symbols we use to represent numbers).

Ok then, does the number that when multiplied by itself equals two have a better way to be represented that has a comprehensible value other than the number that when multiplied by itself that equals two?

 

[Mark44]

Yes, but that just begs the question, how is 1 any different from $$\sqrt{2}$$ ? You can't even draw a line of length '1'.

That's exactly my point.

 

[Mark44]

Ok then, does the number that when multiplied by itself equals two have a better way to be represented that has a comprehensible value other than the number that when multiplied by itself that equals two?

This is a pretty convoluted question, but I think I understand what you are trying to say.

"the number that when multiplied by itself equals two" is $\sqrt{2}$

If that's not comprehensible to you, we can use approximations.

$\sqrt{2} \approx 1.4$
$\sqrt{2} \approx 1.41$
$\sqrt{2} \approx 1.414$
$\sqrt{2} \approx 1.4142$
$\sqrt{2} \approx 1.41421$
$\sqrt{2} \approx 1.414214$

.
.
.
Each successive approximation is closer to $\sqrt{2}$ than the preceding one, but each is still only an approximation. If you square each of these approximations, you'll get a number that's a little under 2, or a little over, but the difference between 2 and the squared value gets smaller and smaller, meaning that the approximations are getting better and better.

 

[WW_III_ANGRY]

This is a pretty convoluted question, but I think I understand what you are trying to say.

"the number that when multiplied by itself equals two" is $\sqrt{2}$

If that's not comprehensible to you, we can use approximations.

$\sqrt{2} \approx 1.4$
$\sqrt{2} \approx 1.41$
$\sqrt{2} \approx 1.414$
$\sqrt{2} \approx 1.4142$
$\sqrt{2} \approx 1.41421$
$\sqrt{2} \approx 1.414214$

.
.
.
Each successive approximation is closer to $\sqrt{2}$ than the preceding one, but each is still only an approximation. If you square each of these approximations, you'll get a number that's a little under 2, or a little over, but the difference between 2 and the squared value gets smaller and smaller, meaning that the approximations are getting better and better.

Thats my issue. It is never exact. For every decimal we expand, when we multiply it by itself the result is not 2, but 1.999...

As I see it there is no value or number more precise than this decimal expansion other than saying the square root of 2 itself or any other form, which essentially is circular in my dilemma. There is no number that when multiplied by itself equals two other than the number that is defined as the number that when multiplied by itself equals two! Am I mad?

 

[WW_III_ANGRY]

That's exactly my point.

Well if we're talking about 1 light year then its a major difference, yes? So vast that it is hard to comprehend how big of a difference that length is.

 

[Mark44]

Thats my issue. It is never exact.

If "it" means the approximations, then no, they are never exact. But $\sqrt{2}$ is the exact value of the number whose square is 2.

For every decimal we expand, when we multiply it by itself the result is not 2, but 1.999...

That's not true. Some of the approximations give values that are larger than 2. For example, 1.414214² = 2.000001238 by my calculator.

As I see it there is no value or number more precise than this decimal expansion other than saying the square root of 2 itself or any other form, which essentially is circular in my dilemma. There is no number that when multiplied by itself equals two other than the number that is defined as the number that when multiplied by itself equals two! Am I mad?

Your error is in thinking that a number's value is its finite decimal representation, which for most numbers is not true. Let's take a simpler (and rational) example - 1/3. If you multiply this number by 3, you get exactly 1. If you want to represent 1/3 by a decimal fraction, no finite-length representation will be exactly equal to 1/3, but the more decimal places you use, the closer you get to 1/3.

If I want to know the number I can multiply by 3 to get 1, it's the number we represent as 1/3. It's not .3, because .3 * 3 = .9. It's not .33, because .33 * 3 = .99, and so on.

 

[WW_III_ANGRY]

If "it" means the approximations, then no, they are never exact. But $\sqrt{2}$ is the exact value of the number whose square is 2.
That's not true. Some of the approximations give values that are larger than 2. For example, 1.414214² = 2.000001238 by my calculator.

Your error is in thinking that a number's value is its finite decimal representation, which for most numbers is not true. Let's take a simpler (and rational) example - 1/3. If you multiply this number by 3, you get exactly 1. If you want to represent 1/3 by a decimal fraction, no finite-length representation will be exactly equal to 1/3, but the more decimal places you use, the closer you get to 1/3.

If I want to know the number I can multiply by 3 to get 1, it's the number we represent as 1/3. It's not .3, because .3 * 3 = .9. It's not .33, because .33 * 3 = .99, and so on.

Mark, thank you for bearing with me :)
Let me continue with my understanding in hope that you can see what I am thinking and where I am wrong:

1/3 is able to be represented precisely in a fraction form. Essentially its a piece divided into three parts. Makes perfect logical sense to me and is comprehensible as well as rational and a rational number. I don't see how the comparison as applicable because my dilemma is with a square root which is reducible to fractions or decimals in rational numbers and is also comprehensible and rational to do so if needed. As I see it everything we know of physically can be divided as fractions or decimal expansions, or what have you. As I conceptualize square roots these too must follow the same path logically or in the case of irrational numbers these aren't applicable.

I understand geometrically that 1/3 is more desirable than .333, however as stated above this is able to be comprehended conceptually and physically. I understand the square root of 2 is able to be utilized to find the hypotenuse of a right triangle, however that is an incomprehensible value to me as well as it is unable to be expressed in the physical, or conceptually through decimal expansion or fraction. Which is why I see it as irrational in the whole sense of the word :)

 

[WW_III_ANGRY]

If "it" means the approximations, then no, they are never exact. But $\sqrt{2}$ is the exact value of the number whose square is 2.
That's not true. Some of the approximations give values that are larger than 2. For example, 1.414214² = 2.000001238 by my calculator.
.

Actually the square root is 1.414213 yes? Which I still show as 1.9999... (with some string of other integers dependent upon how far out we go at the tail end)

 

[Mark44]

Actually the square root is 1.414213 yes?

No. I rounded up, and you rounded down. sqrt(2) is between 1.414213 and 1.414214.

Which I still show as 1.9999... (with some string of other integers dependent upon how far out we go at the tail end)

 

[WW_III_ANGRY]

No. I rounded up, and you rounded down. sqrt(2) is between 1.414213 and 1.414214.

Oh I see. I didn't wish to round, I terminated the sequence without rounding.

 

[JSuarez]

however that is an incomprehensible value to me as well as it is unable to be expressed in the physical, or conceptually through decimal expansion or fraction. Which is why I see it as irrational in the whole sense of the word :)

But why should these be the only possible ways of "comprehending" that particular root? Consider the sequence:

$$x_0=1$$
$$x_{n+1}=\frac{2-{x^2}_n}{2x_n}$$

It has a limit, equal to the positive root of $x^2-2$ and both these facts may be proven independently; moreover, if you multiply it termwise by itself, you'll get again a convergent sequence, whose limit is 4.

 

[jbunniii]

As I understand it also, finding the numerical value in decimals is the most precise method to evaluate the square root of 2. If there is another more precise method can you please let me know? Thanks,

I think you are conflating two different issues.

Not every real number has a decimal expansion with finitely many digits. In fact, most do not.

But this is not unique to irrational numbers. There are many rational numbers that do not have terminating decimal expansions. 1/3 is an example.

That $\sqrt{2}$ is irrational means precisely that it cannot be expressed as a ratio of two integers: there are no $m,n \in \mathbb{Z}$ such that

$$\left(\frac{m}{n}\right)^2 = 2$$.

This shows that 2 does not have a square root in the rational field $\mathbb{Q}$. The existence of such gaps in $\mathbb{Q}$ is the motivating factor behind the construction of the larger field $\mathbb{R}$ (real numbers), which does not have such gaps.

There are several well-defined ways to construct $\mathbb{R}$ from $\mathbb{Q}$. One standard method uses Cauchy sequences, and another one uses Dedekind cuts. Just about any real analysis book should contain at least one of these constructions. Some calculus books do, too. (E.g., Spivak.)