284 = 4^4 + 4! + 4
And someone please correct me if I am so horrendously wrong with this next one since I am really sick atm and brain is wanting to just sleep all day...
285 = [ 4!(4!) / √4 ] - {√4}?
If I'm wrong, forgive me for wasting everyone's time, as this is the first time I've encountered ? and ! in a VERY LONG time. [And it not be in the middle of some spanish instruction manual accidentally shipped out instead of an english one]
Mr. Kelly R. Flewin
Mr. Kelly R. Flewin
Just another random gamer
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<OmnipotentEntity> How do you people get bored in the span of 10 seconds? Worst ADD ever.
wait, school might be past from a long time, but there's something fishy there.
since when :
√4 = 2 ?!?!
that's not true, √4 = 2 or -2
( -2 * -2 = 4 so -2 is a root)
one should correct :
|√4| = 2
it seems so basic that I'm even not sure...
I never sleep, 'cause sleep is the cousin of death - NAS
That's aboslutely false SXL...
you're right on the fact that (-2)² = 4 and 2² = 4...
THOUGH √ x is ALWAYS a positive number...
The best proof of that is graphic of the function : f(x) = √x this function is strictly "ascending" and always positive...
Click here to see the graphic of this fonction
Oh come on, it's commonly understood that √x usually means the positive root, even if two exist. Don't be an ass.
Xaphan, techincally he's right. The graph should look like the function y = x^2 flipped over the line y = x, so it should be a sideways parabola.
Anyway, after this, who's up for the One One challenge?
Anyway, after this, who's up for the One One challenge?
LOL... if anyone try it , he will surely have a very VERY harsh time... HAHAHA :) (if you try it, you'll be as ridiculous as if you try to lick your elbow...)
Anyway maybe a 2 2's challenge could be possible (I did not say interesting...)
The only reason most calculators dont show -2 on the graph of the square root of x when x=2 is because they can only graph 1 x for every y.
It's not only a calculator problem... It's the way the word "function" has been defined... functions must be bijective (i don't know if it's the good english word) this means one X must give one and only Y...
293 = [ ( 4 ! x 4 ! ) + 4 ? ] / √ 4
(if you try it, you'll be as ridiculous as if you try to lick your elbow...)
I can do that. Very flexible arms. picture showing my arms(not me licking elbow though)
since when :
√4 = 2 ?!?!
that's not true, √4 = 2 or -2
( -2 * -2 = 4 so -2 is a root)
one should correct :
|√4| = 2
He is completly correct. Don't call him an ass either. And don't call me an ass for agreeing with him.
Try doing Quadratics when you need to find both zero's(roots) in able to find Vertix's, length of a throw, etc...
It's not only a calculator problem... It's the way the word "function" has been defined... functions must be bijective (i don't know if it's the good english word) this means one X must give one and only Y...
The negative value for X would give the same value for Y as the Positive would.
edit: only if its a Quadratic.
That's not the same problem... finding roots for that will give you 2 solutions as (-x)² = x²...
The fact is that √ x is positive as it has been defined so... √ x is always positive even if it's confusing...
This pretty more understandable with those two properties :
√ x = x^(1/2)
√ x² = |x|
x is a positive number (or you'll get a complex result) so x^(1/2) is also positive... Check it if you do not trust me...
294 = [ 4 ! x ( √ 4 ) ? ] x 4 + ( ( √ 4 ) ? ) !
Joined: 4/20/2005
Posts: 2161
Location: Norrköping, Sweden
Theoretically, you could go on forever, since there is no number that is to high to write with 4 4's. With just one 4 you could get amazingly high numbers, for example, ((4?)?)! = 1,27e+73, but I doubt we will go that far with this :p
379 = (4!)? + (4?)? + (√4*√4)!
380 = (4!)? +(4+4)*4?
Did you try one 1 Randil ???? :)
I'm wondering how far we will be able to go (without getting bored) with the 4 4s challenge...
try the 2 2s challenge :)