/sci/ - Science & Math

[Anonymous No.7048105]

Quick question for you smart /sci/ducks:

Is it possible to divide an integer by the square root of 2 and still get an integer number?

[Anonymous No.7048106]

No

[Anonymous No.7048109]

no dawg

[Anonymous No.7048113]

>>7048106
Is there any way to divide anything by the square root of 2 (that isn't the square root of 2) and get an integer?

[Anonymous No.7048115]

nice homework thread faggot

[Anonymous No.7048116]

>>7048113

Of course. 2sqrt(2). Divide by sqrt(2). Get 2.

[Anonymous No.7048117]

>>7048115
I do agree !

>>7048113
I think yes. Have some ideas, but will let you do your homeworks. :-)

[Anonymous No.7048120]

>>7048115
It's not a homework thread.

Now, I know that an irrational number cannot be expressed as the ratio between two integers, so if I have something like: <div class="math">\sqrt{2} = \frac{x}{n}</div>
where n is an integer, is it then always true that x is irrational? Or is it simply some sort of decimal number?

[Anonymous No.7048127]

>>7048105
so are all of you just ignoring the case where 0 is the integer or no

[Anonymous No.7048131]

>>7048120
Damn, man, that's real stuff.
I throw in the towel, I'm in humanities now.

[Anonymous No.7048135]

>>7048120

Any decimal of finite length is rational. eg 3.1415 = 31415/10000

[Anonymous No.7048136]

>>7048109
isn't that wrong because not every number is a complex number?

[Anonymous No.7048146]

>>7048135
Yeah, I worked it out slightly differently, but x will always be irrational.

[Anonymous No.7048148]

>>7048136
every number is a complex number. given any number n, it can be expressed in complex notation as n + 0*i. there's just no imaginary part.

[Anonymous No.7048159]

So is it possible to add two irrational numbers and have the result be a rational number?

[Anonymous No.7048162]

>>7048159

yes

[Anonymous No.7048163]

>>7048105
No.
Proof:
Assume <span class="math">\frac{n}{\sqrt{2}} = m\\ then\;we\; have: n = \sqrt{2} m, or \\ \sqrt{2} = \frac{n}{m}[/spoiler] .
This fails, because we know <span class="math">\sqrt{2}[/spoiler] to be irrational.

[Anonymous No.7048165]

>>7048163
<span class="math"> \frac{n}{\sqrt{2}} = m \Rightarrow n = \sqrt{2} m \Rightarrow \sqrt{2} = \frac{n}{m}[/spoiler]

[Anonymous No.7048214]

>>7048109
what's the difference between a whole and an integer number?

[Anonymous No.7048215]

>>7048214
Whole number is a non-negative integer?

[Anonymous No.7048219]

>>7048109
I think I'm being silly bu what is an example of a non rational, non irrational real number?

[Anonymous No.7048302]

>>7048215
But that's what natural numbers are.

[Anonymous No.7048315]

>>7048302
Zero isn't a natural number.

[Anonymous No.7048374]

>>7048219
There are none. A real number is either rational or irrational.

[Anonymous No.7048381]

>>7048315
So whole numbers is literally just N U {0}?

The diagram's retarded anyway, "irrationals" shouldn't be its own subset just the remainder of the real numbers outside the set of rationals.

[Anonymous No.7048403]

0, you dumbfucks.

Z is a ring, by definition it has to have an additive identity element. That's 0.

0 divided by anything is 0.

It is, however, the only integer for which this is true.

[Anonymous No.7048899]

>>7048159
given irrational number n, and considering the fact that (1 - n) is irrational, by the following proof

>Assume that for any irrational number n, (1 - n) = x where x is rational.
>1 - n = x <=> 1 - x = n
>we know 1 and x are rational, so any calculation concerning only them two will be rational. however, n is irrational. thus our claim is false and 1 - n must be an irrational number

right, since n is irrational and (1 - n) is irrational, you can add n and (1 - n) together to get 1, which is a rational number. thus, it is possible to add two irrational numbers and have the result be a rational number.

[Anonymous No.7048906]

>>7048403
already given here
>>7048127

[Anonymous No.7050787]

Of course you can do that, it's quite easy

Here you are:

42426406871100000000000000000000/1.41421356237
30000000000000000000000000000000.000000000000000000000000000000

.

[Anonymous No.7050790]

>>7048105
yes

0/√2=0

[Anonymous No.7050833]

>>7050790
You cannot divide zero by a number.

[Anonymous No.7051082]

>>7050833
:^)

[Anonymous No.7051477]

>>7048105
no.
2^(1/2) is irrational

[Anonymous No.7051478]

>>7048105
0/sqrt(2) = 0

[Anonymous No.7051493]

>>7048105
Is /o/ really this fucking retarded? This is first day analysis I

[Anonymous No.7051494]

>>7051493

[Anonymous No.7051556]

>>7048374
then whats in the green area thats outside the red and blue area?

[Anonymous No.7052137]

>>7048109
>set of real numbers is larger than the set of non-real complex numbers
>set of rational numbers is larger than the set of irrational numbers
>set of non-natural whole numbers({0}) is larger than the set of natural numbers
>there are real numbers which are neither rational nor irrational

[Anonymous No.7052141]

>>7048148
What about uncomputable numbers?

[Anonymous No.7052175]

>>7048105
Its not possible do divide by a number that doesn't exist

[Anonymous No.7052177]

>>7048105
When diving a rational and irrational number the result is always irrational.

[Anonymous No.7052189]

>>7052137
>>set of real numbers is larger than the set of non-real complex numbers
>>set of rational numbers is larger than the set of irrational numbers
>>set of non-natural whole numbers({0}) is larger than the set of natural numbers
Nigger you need to go back to 5th grade and learn how to read a Venn diagram. How big the circle is does not represent how large the set is in relative terms, that is in no way shape or form ever implied in the nature of such a chart, if you are attempting to make a chart similar to a Venn diagram that actually sizes the regions relative to the size of the sets they represent you need to clarify that you are doing so because there is no reason to assume that you are.

In some situations this might be reasonable but this is quite clearly not one of those cases, the reals are an infinitesimal subset of the complex numbers, the rationals are an infinitesimal subset of the reals so it is impossible to draw a Venn diagram showing all 3 in proper proportionality, in both cases the smaller set wouldn't even get a single pixel, you'd just have a disjoint series of circle graphs telling you nothing. Absolutely retarded.

>>set of non-natural whole numbers({0}) is larger than the set of natural numbers
>>there are real numbers which are neither rational nor irrational
Well you are right that it is a completely retarded graphic for the latter and for even including N U {0} as a separate set from N but you still do not know how to read an elementary school level chart.

[Anonymous No.7052195]

>>7052189
Seriously, all such a chart says is<div class="math">\mathbf{N} \subset \mathbf{Z} \subset \mathbf{Q} \subset \mathbf{A} \subset \mathbf{R} \subset \mathbf{C}\,.</div>Nothing else.

[Anonymous No.7052196]

stealing bread

is it possible that i have learned linear algebra through previous math courses I have taken?

currently taking differential equations that has linear algebra as an optional prereq. I have taken

in HS
>alg 1,2
>precal
>calc AB
>geometry

in uni
>calc 1,2,3 (differential, integral, and multivariable)

so yea do i know the important stuff from linear algebra?

[Anonymous No.7052201]

>>7052196
You do not know linear algebra.
Definitely take linear algebra, its better than diff eq anyway

[Anonymous No.7052243]

>>7052137
I don't like this guy, but I approve of this image.

[Anonymous No.7052286]

>>7052201
its too late for me to register for a class without instructor approval

how fucked am i for DE?

[Anonymous No.7052336]

>>7052286
Not fucked at all, all you need from linear algebra is bowing how to row reduce a matrix and find eigen vectors, you can learn this really quickly by yourself.

[Anonymous No.7052396]

>>7050787
>It's real to 12 significant figures
nah man