Can you give me an example of an irrational number, NOT including Pi?

Forty twelve?

the square root of 2.

e.

The golden ratio. Damn, I can’t make a phi in Fluther.

square root of 3, 5, 6 7 and 8 ? I’m bad at math.

@cazzie I’m assuming you are listing one third as irrational because it is repeating in decimal form. However, an irrational number needs to be a real number that is not terminating or repeatable when written in decimal form. If that isn’t your reasoning, I apologize for assuming, but one third is still a rational number.

@amujinx I corrected myself already, thank you. I realised as soon as I hit Answer that it was wrong. That will teach me to try to do math before my second cup of coffee in the morning.

@cazzie Yes, I saw. However, I didn’t notice when I submitted it, and now it’s too late edit it. Sorry.

π-3, π+3, π×3, π/3, and √7.

Eleventyhundredteen

Anything that cannot be expressed in fraction form. If it’s a fraction, it’s rational.

The square root of any number that isn’t a perfect square is irrational. Sqrt(3) is irrational, sqrt(4) isn’t (it is 2 (or -2), because 2^2 = 4).

Square roots of negative numbers are always irrational.

Thanks guys. Every time I see the abbreviation “sqrt” for “square root” I think of….well, diarrhea. Well! I do!
I’ve gone as far as college algebra, and I don’t recall having to deal with any irrational numbers other than π (thanks for the symbol @ratboy)! At what level do irrational numbers start becoming a factor?

@CWOTUS I might be being overly pendantic, but aren’t irrational numbers a subset of real numbers? And square roots of negative numbers are imaginary.

@Mariah Aren’t real numbers rather a subset of irrational numbers?

You may be correct, @Mariah. I’m a long way removed from my instruction about such things, and to this point the instruction has been perfectly useless.

@flutherother No. There are no irrational numbers that end in repeating numbers.

Correction: irrational numbers are a subset of real numbers. You are right @Mariah

Pi belongs to a subset of irrational numbers known as transcendental numbers. The square root of 2, on the other hand, is algebraic rather than transcendental because it can be expressed as the root of an algebraic (polynomial) equation: x^2 – 2 = 0. There is no such equation for pi.

ln 2, ln 3, ln 4, ln 5, ln 6…

@mattbrowne Yes, those are all transcendental numbers, too, along with e & various notable constants. Curiously e^pi is known to be transcendental, while pi^e is “not proven to be transcendental, but generally believed to be by mathematicians.” [ Ref ]

The proof is left to the reader as an exercise, lol

@gasman – I wasn’t aware of that. Thanks! Well, using a calculator pi^e =

22.459157718361045473427152204543735027589315133996692249203 002554066926040399117912318519752727143031531450073148896372 716654162727200036841245878483825780197399927516270911185238 671352940834892162337692496730536751662599601668725547775888 060873742920118171661161372246197209044896331314599273279140 914840576764889753784885344102006492593490357594763469165286 294007847395407755298019829026131224029511379990688652442931 146335939285716073329541491532530137722767555183068793043622 842319088286797578297967264718164000704357181058364260641458 021223969069744749885161613318405106216364559561084133094289 92831263755783615…

So, if it’s not transcendental, does this mean there’s some rational number q with sq(q)=pi^e ?

I love pie…. Ever think how cool it is that pie is round? I like my pi to have some pie involved.

I also love rectangular pies with a nice non-transcendental diagonal, like

http://www.hockeyweb.de/pictures/donauwelle2-07-do.jpg

@mattbrowne If it’s not transcendental then it’s algebraic, which means it’s the root of an equation of the form:
ax^n + bx^(n-1) + cx^(n-2) +...+ ux + v = 0
where all the coefficients a, b, etc are integers and n may be large but must be finite.

I didn’t mean to imply that all algebraic numbers are the square root of a rational. The cube root of 2, for example, is not the square root of any rational number even though it’s algebraic because it satisfies x^3 – 2 = 0.

Cantor showed that the algebraics are countable just like the integers and rational numbers. Transcendental (non-algebraic) numbers, on the other hand, are uncountable like the set of real numbers itself. So in a sense, there are “more” transcendental numbers than there are algebraic numbers, even though they’re both infinite sets.

@cazzie ”...pie is round.” When it comes to figuring the area of a circle, “pie are square”!

@gasman I don’t want to figure out the area of my pie… I just want to eat it. You can’t count your pie and eat it too.