How can sqrt(2) be considered a real number when it has no solution? There is no point on a number line that is sqrt(2), as soon as you define a point as sqrt(2) you have given it finite decimals places making it not sqrt(2).
How can sqrt(2) be considered a real number when it has no solution? There is no point on a number line that is sqrt(2)...
Gabriel Nelson
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Christopher Sanchez
No
Yes
Maybe
Robert Peterson
Wilderburger pls
Jonathan Parker
You're getting rational/irrational and real/complex mixed up.
Do high school algebra again please and spare us your stupidity.
Julian Nelson
Your statement is wrong.
Jeremiah Reed
[math]\sqrt{2}[/math] is formally the dedekind cut of all [math]x \in \mathbb{Q} [/math] such that [math]x^2 < 2 [/math].
Learn the definition of [math]\mathbb{R}[/math]; you clearly don't know what a real number is.
Henry Barnes
>How can sqrt(2) be considered a real number when it has no solution
You can express sqrt(2) as infinite series. If you don't believe in infinity you're a fag.
>define a point as sqrt(2)
>you have given it finite decimals places making it not sqrt(2)
What? That is only true for discrete spaces. R is complete.
Connor Brooks
It is a non repeating sequence, which means it is not a series, and there is no concept of convergence here.
Gabriel Davis
yes, there is
do you know what a real number is? its either a cauchy sequence or a dedekind cut
Julian Carter
I know what "real numbers" are meant to be little boy. Why don't you think for yourself instead of parotting definitions thinking you're smart?
Xavier Wright
>you can express infinity. Would you mind writing down an infinitely long number for me then?
Jaxon Perez
what do you mean by 'solution' ?
you perform the square root function on the number 2 and get a number that is the square root of two .which happens to be a real number .
>as soon as you define a point as sqrt(2) you have given it finite decimals places making it not sqrt(2).
having infinite decimal places dosnt make it not the square root of two .for example you have 0.999999.... which equals 1 exactly .
but you shouldn't be using decimals to represent the square root of 2 ,there no point .
Brayden Jackson
you showed in all your posts you dont have the faintest idea of what a real is
Evan Clark
and what number is the square root of 2 then? There's no number.
Tyler Russell
of course there is a number, it's just infinitely long
obviously you can't express every number physically .. there's a smallest length
is op autistic
Adrian Perez
By the definition of real numbers OP is a retard
Easton Perry
>It's just infinitely long
Such things don't exist and to treat such things as real numbers is absurd
Carson White
then define existence; of course they don't "exist", but unless you're retarded you can conceptualize them and calculate them to whatever significant digit that gets your jimmies wet
Luke Price
It doesn't exist meaning that there will never be a solution because as you said it has infinite decimals. You can't come to the end of infinity.
Ryan Ross
youtu.be
enlighten yourself
Alexander Sullivan
>How can 1/3 be considered a real number when it has no solution? There is no point on a number line that is 1/3, as soon as you define a point as 1/3 you have given it finite decimals places making it not 1/3.
How is writing "1/3" for a number different from using "sqrt(2)" for another? Neither are decimal representations.
William Walker
Then don't bother with decimal expansions of numbers. Problem solved.
Joshua Ortiz
Significant figures.