Not true. Gryle stated the following: X = 0.9r 10X = 9.9r 9X = 9 X = 1 1 = X = 0.9r If we replace 0.9r by 0.7r, we get: X = 0.7r 10X = 7.7r 9X = 7 (remember, this step is no division, it is merely subtraction, -X on both sides) X = 0.7r It really only works with 0.9r. About the bases, we use a base-12 system every day when we indicate the time. In fact, hours lasting 60 minutes of 60 seconds each would rather suggest a base-60 system.
So there are bases that have their uses, funny. Now find a use for base 7 and base 43 You are correct Roel it does not lead to 1, my bad. Instead it leads to 9/7 (if we divide by 7 to make the left side equal 1) which is more then 1 even further proving it is incorrect.
Eh? 9X = 7 means that X = 7/9, not X = 9/7. That's why X, in this equasion, is indeed 0.7r (which is less than one) and nothing else.
* ∞ I blame my physics/math professor, that guy is on such a high level, even his colleagues get confused by his rushing through problems (Try as he may he can't step down, which sometimes makes him a lousy teacher) Back to topic. A simple, no tricks problem; x+x=x*x=x^x x=? (multiple solutions) As for the a=b problem, the moment you see a+b=b something has gone wrong because x can never =x+something. The real solution eludes me at this late hour though.
One interesting point is, that the logical consequence of the proofs of 1 = 0,9r means, that all real numbers will have two values, since the same reasoning can be applied to all other real numbers as well. Stix Aside from 2, which other numbers will fit into the equation?
Right, I'm with you now. No, reading through your last few posts it's clear you seem to have some funny ideas about recurring decimals, there is no "carried 1" on the end the carry only exists while the operation is being performed it dosen't exist in the answer, 0.3r contains exactly one 0 and an infinite number of 3s and it is equal to 1/3. No it shouldn't be, you realise that 8.9r1 is complete gibberish right? If it's infinite it has no end to put the 1 on, if you do have a 1 on the end then it's finite and therefore not recurring. And going by how people voted in that thread 55.88% agree with the proof I supplied earlier. Here, read this 0.999... it has a whole slew of other proofs that may address some of your concerns, although given what you've posted so far I wouldn't hold my breath. No, they will have the same value, the same number, but 2 different representations as I said earlier the decimal representations for some numbers aren't unique. Specifically any Real number with a terminating decimal representation.
Actually, according to my calculator 0^0 isn't possible. Either way, the next time I pose this riddle I'll conceit that there is only 1 solution.
but 0^n = 0 so 0^0 is a special case (undefined) One of my students once suggested that since x^0 is 1 and 0^n is 0, 0^0 should be (splitting the difference) = 1/2 then again, when faced with sin x / n, he cancelled the n's and got 6!!!!!
...I DO THINK CHRISTIANS oops sry caps ...precise logic clearly defines the women /evil debate once and for all ....
maths Hello Woody, re your question about the spinning mouse.... the answer is..... "Because the higher it gets, the fewer".
Not quite: First we assert, that girls require time and money: Girls = Time x Money should be time + money As we all know, time equals money: Girls = Money x 2 Money, as we know, is the root of all evil: Money = sqrt(Evil) Girls = Money x 2 Therefore Girls = 2 x sqrt(Evil) transpose: Girls/2 = sqrt(Evil) square both sides (1/2 x 1/2 is 1/4) = girls are only 1/4 evil
Tut tut Roxy... Time AND money AND etc... addition, not multiplication. In addition to which the penultimate statement is demonstrably wrong. It should read "men ARE worse than evil". PS the second sentence also amuses.... "sate" men? That's a woman's job: leave the men satiated!