This is one of those math puzzles I can never figure out and always slap my head once they’ve been explained.
Take 2 numbers a and b, but a =/= b
Take the mean to be x
x= (a+b)/2
2x=(a+b)
2x(a-b)=(a+b)(a-b)
2xa-2xb=a[sup]2[/sup]-b[sup]2[/sup]
a[sup]2[/sup]-2xa=b[sup]2[/sup]-2xb
a[sup]2[/sup]-2xa+x[sup]2[/sup]=b[sup]2[/sup]-2xb+x[sup]2[/sup]
(a-x)[sup]2[/sup]=(b-x)[sup]2[/sup]
a-x=b-x
a=b
But we defined a=/=b
What am I missing?
Your error occurs in this line. There are two roots to consider, the positive and negative, so…
either a-x = b - x … or … a - x = x - b
Because a /= b, we know that
a - x = x - b … or in other words x is the arithmetic mean of a and b (as defined).
nm
Yup. In these puzzles, typically one of two things happens: division by zero, or taking square roots and ignoring the existence of two possible roots. As hdc_bst found, in this case it’s the latter here.
Figured as much. Thanks folks.
I’ve also seen them based on omitting a constant of integration. IIRC, it made use of logs, since for most integrals, leaving off the constant will give you the integral from 0 to your point, but for the integral of 1/x, you find the integral from 1 to your point (since the integral from 0 diverges).
Also, this line is incorrect. So to move the 2xa over to the other side, you subtract it and that works out.
But to move the 2xb to the other side you ADD it, which means it’s b[sup]2[/sup]+2xb
Are you sure about that? :dubious:
What would be incorrect?
To move -2xb to the other side, you have to add it on both ends.
Actually, what would happen is you’d get -b[sup]2[/sup]+2xb, which you could convert to b[sup]2[/sup]-2xb but only if you put a -1 on the other side as well. Either way it doesn’t work.
You’re wrong. He’s added b[sup]2[/sup] to both sides, and left the -2xb where it was…
2xa-2xb=a[sup]2[/sup]-b[sup]2[/sup] <— line from the OP
add -2xa to both sides, gives
-2xb=a[sup]2[/sup]-b[sup]2[/sup]-2xa
re-arrange terms on right side, gives
-2xb=a[sup]2[/sup]-2xa-b[sup]2[/sup]
add b[sup]2[/sup] to both sides, gives
b[sup]2[/sup]-2xb=a[sup]2[/sup]-2xa
swap sides, gives
a[sup]2[/sup]-2xa=b[sup]2[/sup]-2xb <— line from the OP
A trivial observation on hdc’s solution: If X is the mean of A and B and A!=B, X must be halfway between A and B, i.e. smaller than one of them but bigger than the other, and by the same amount each time.
Therefore, either: A - X = X - B (and B - X = X - A).
Off to The Game Room.
Colibri
General Questions Moderator
Oh
Well how about that?
In general, an easy way to find a step where things go wrong: Plug in actual numbers for a and b and walk through the lines till you find the first equality that isn’t actually true with those specific values. If you did this, it would readily have been apparent that the step moving from (a - x)^2 = (b- x)^2 to a - x = b - x was the source of the problem.