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Square root of 2

by MonkeyKid
Tags: root, square
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MonkeyKid
#19
Feb28-14, 04:59 PM
P: 25
Quote Quote by Mark44 View Post
There's no such thing as a two dimensional length. Length is one dimensional measure. On the other hand, area is a two dimensional measure, and volume is a three dimensional measure.
Even if you had a ruler marked in units of ##\sqrt{2}##, you wouldn't be able to tell if what you were measuring was actually ##\sqrt{2}## in length. Measuring things is inherently inaccurate.
you are right about the dimensions of length. Sorry, my mistake. About measuring, I was talking more of a hypothetical measuring with arbitrary precision. I thought that was implied. And what I meant was that even if you could measure things with perfect precision, unless you used a unit based on ##\sqrt{2}## you could never measure the whole segment
PeroK
#20
Feb28-14, 05:45 PM
P: 377
You can look at this a different way. We know that 1.4 < √2 and 1.5 > √2. And we can keep going with every greater mathematical precision:

[tex]1.41421356^2 < 2 \ and \ 1.41421357^2 > 2[/tex]

And so on. And, we can make these two numbers either side of √2 as close as we want. So, the question is simple: is there an "irrational" number in there that is equal to √2 or not?

Suppose not, then the number line has gaps in it. And, it would get quite difficult to do maths if there are gaps where you'd like numbers to be. For example, if you draw the graph of

[tex]y = x^2-2[/tex]
This clearly crosses the x-axis twice (at -√2 and +√2). But, if there are gaps where these numbers should be, then the graph crosses the x-axis between numbers, so there is no solution to the equation. The graph sort of ghosts thru the x-axis missing all the numbers.

So, an axiom of mathematics is that there are no gaps. This is called the "completeness" axiom.

It might be quite interesting to see how far you could go without the completeness axiom. But, intuitively, I don't like the idea of a curve crossing the x-axis where there is no number! Surely, wherever the curve crosses the x-axis, there must be a number there?
MonkeyKid
#21
Mar2-14, 05:24 AM
P: 25
Quote Quote by PeroK View Post
You can look at this a different way. We know that 1.4 < √2 and 1.5 > √2. And we can keep going with every greater mathematical precision:

[tex]1.41421356^2 < 2 \ and \ 1.41421357^2 > 2[/tex]

And so on. And, we can make these two numbers either side of √2 as close as we want. So, the question is simple: is there an "irrational" number in there that is equal to √2 or not?

Suppose not, then the number line has gaps in it. And, it would get quite difficult to do maths if there are gaps where you'd like numbers to be. For example, if you draw the graph of

[tex]y = x^2-2[/tex]
This clearly crosses the x-axis twice (at -√2 and +√2). But, if there are gaps where these numbers should be, then the graph crosses the x-axis between numbers, so there is no solution to the equation. The graph sort of ghosts thru the x-axis missing all the numbers.

So, an axiom of mathematics is that there are no gaps. This is called the "completeness" axiom.

It might be quite interesting to see how far you could go without the completeness axiom. But, intuitively, I don't like the idea of a curve crossing the x-axis where there is no number! Surely, wherever the curve crosses the x-axis, there must be a number there?
Because of these numbers being necessary, I thought they were made up, so math could work, even though there would be no exact point whose distance from zero in the real number's line would be the [itex]\sqrt{2}[/itex], since I had no proof of it's existance. So, when I first asked the post, I thought you could get arbitrarily close to [itex]\sqrt{2}[/itex] but there was no actual [itex]\sqrt{2}[/itex].

After the geometrical demonstrations I was shown, I was convinced [itex]\sqrt{2}[/itex] indeed exists, because it is possible to draw a line segment with a length of [itex]\sqrt{2}[/itex]. Then if you take that line segment and put it's begining at 0 in the real numbers' line, it's end would be the exact point of the line that has a distance equal to [itex]\sqrt{2}[/itex] from 0, in other words, that point is where [itex]\sqrt{2}[/itex] lies in the real number's line.

Geometrically, it's easy to picture. Using just numbers, not so much, at least for me. I suppose my mind is limited to things more concrete like lines and points and circles, and is bad at conceiving purely abstract stuff.
MonkeyKid
#22
Mar2-14, 05:32 AM
P: 25
Quote Quote by gopher_p View Post
Perpendiculars are constructible with a compass and straight edge, so you don't [i]need[\i] to assume the existence of right triangles in this setting. Note that you can't construct all right triangles since not all lengths or angles are constructible.

There are other kinds of constructions that are more "lenient" than compass and straightedge in terms of what is assumed/allowed (https://en.wikipedia.org/wiki/Compas..._constructions).
I know this kind of strays from the original subject, but how can I, without the notion of the right triangle and it's properties, and armed only with the notion of perpendicular angles, know that the line that connects two line segments that are perpendicular to each other has the square of it's length equal to the sum of the square of the lengths of the two line segments? (I don't promisse to understand the explanation, but I will read it diligently, and try my best, hopefully learning new things along the way)
micromass
#23
Mar2-14, 09:03 AM
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Quote Quote by MonkeyKid View Post
I know this kind of strays from the original subject, but how can I, without the notion of the right triangle and it's properties, and armed only with the notion of perpendicular angles, know that the line that connects two line segments that are perpendicular to each other has the square of it's length equal to the sum of the square of the lengths of the two line segments? (I don't promisse to understand the explanation, but I will read it diligently, and try my best, hopefully learning new things along the way)
Are you looking for a proof of the Pythagorean theorem?
MonkeyKid
#24
Mar2-14, 09:21 AM
P: 25
Quote Quote by micromass View Post
Are you looking for a proof of the Pythagorean theorem?
Yes, one that doesn't assume the existance of triangles or any other geometrical entities other than lines, points and circles. The only one I was taught which I can't quite recall uses the area of four triangles and the area of the square to prove it.
micromass
#25
Mar2-14, 09:27 AM
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Quote Quote by MonkeyKid View Post
Yes, one that doesn't assume the existance of triangles or any other geometrical entities other than lines, points and circles. The only one I was taught which I can't quite recall uses the area of four triangles and the area of the square to prove it.
Are you ok with an axiom system mentioning only lines and points? The axiom system will then develop notions of triangles and squares to help prove the Pythagorean theorem. Or do you want a text that only talks about lines, points and circles and nothing else (in which case I doubt you'll find anything useful).

Anyway, a very good book is http://www.amazon.com/Geometry-Eucli.../dp/0387986502
In the second chapter he starts with constructing an axiom system using the undefined notions of "line", "point", "between" and "congruent". It might just be what you're looking for. The book is however a bit to the difficult side.
MonkeyKid
#26
Mar2-14, 10:44 AM
P: 25
Quote Quote by micromass View Post
Are you ok with an axiom system mentioning only lines and points? The axiom system will then develop notions of triangles and squares to help prove the Pythagorean theorem. Or do you want a text that only talks about lines, points and circles and nothing else (in which case I doubt you'll find anything useful).

Anyway, a very good book is http://www.amazon.com/Geometry-Eucli.../dp/0387986502
In the second chapter he starts with constructing an axiom system using the undefined notions of "line", "point", "between" and "congruent". It might just be what you're looking for. The book is however a bit to the difficult side.
Quote Quote by gopher_p View Post
I think the constructibility (https://en.wikipedia.org/wiki/Constructible_number) of ##\sqrt{2}## is the most compelling and understandable argument for why it should exist, especially given the OP's apparent beliefs regarding the existence of circles and lines.
The reason why I mentioned the possible need for a proof that does not involves triangles is because this article above uses something called constructible numbers to prove that [itex]\sqrt{2}[/itex] is a number. Using compass and a straightedge, it draws a square with side 1 and then concludes that the diagonal of the square is equal to [itex]\sqrt{2}[/itex]. Well, unless this conclusion comes from something other than the knowledge of the pythagoras' theorem, I don't see the need for all the compass and straightedge operations. All you have to do is declare something like this: "say we have a right triangle where both catheti have a length of 1; according to the pythagoras' theorem the hypotenuse will have a length of [itex]\sqrt{2}[/itex]". So, if the proof uses the pythagoras' theorem I can't see the point of all that hard work in using compass and straightedge, unless there's a way to prove the pythagora's theorem without taking right triangles into consideration.


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