The final question on my alg2 exam last year.

[spyke252]

Re: The final question on my alg2 exam last year.

Hm...
There aren't any visible problems with that proof.
However, the 4=2 is unnecessary, as if -1=1, then any number is equal to any other number.
But I assumed you added 3 to both sides to get the 2=1 answer.

[Afrobean]

Re: The final question on my alg2 exam last year.

Oh I see the problem now.

Because of the square root, it's really plus or minus.

It's sort of like saying this:

±1 = ±1
1 = -1

Which is obviously wrong.

[Kit-]

Re: The final question on my alg2 exam last year.

Actually, square roots are by definition positive. But you're getting close.

[Afrobean]

Re: The final question on my alg2 exam last year.

Actually, square roots are by definition positive. But you're getting close.

rofl

So I guess -1 squared isn't 1.

You might simplify things by saying that square roots are always the absolute value, but that's not entirely mathematically correct.

[Kit-]

Re: The final question on my alg2 exam last year.

Solutions to x^2-a=0 are not by definition positive.
Solutions to x=a^(1/2) are by definition positive.

[Reach]

Re: The final question on my alg2 exam last year.

There is a contradiction from lines 3 to 4.

Root (1/-1) does not equal root(1)/root(-1), hence the rest is all incorrect.

i^2= -1

-1=-1

[spyke252]

Re: The final question on my alg2 exam last year.

Root (1/-1) does not equal root(1)/root(-1), hence the rest is all incorrect.

I thought that too, but it is true.

[Reach]

Re: The final question on my alg2 exam last year.

Uh no it isn't. You can't take the root of the denominator there unless it's positive because the result does not exist.

[spyke252]

Re: The final question on my alg2 exam last year.

EDIT, crap. reach is right.

The calculation rule

(a^.5) * (b^.5) = (a*b)^.5

is only valid for real, non-negative values of a and b

[chickendude]

Re: The final question on my alg2 exam last year.

Proof: Everyone in the world is the same age

Let n = number of people in the world (a group)
Let s(n) = the maximum number of different ages existing in any group of n

s(1) = 1, (with one person in the world, everyone in the world is the same age)

assume s(k) = 1:
by showing that s(k+1) = 1, we show thateveryone in the world has the same age by induction

Take two people in the group of k+1. Call these people Al and Bob.
Take the group of everyone except Al. This is a group of k people. By the assumption, everyone in this group has the same age.
Take the group of everyone except Bob. This is a group of k people. By the assumption, everyone in this group has the same age.

Take a third person, Carol. Carol is the same age as Bob and is also the same age as Al by the two statements above, therefore Bob and Al are the same age.

Since Al and Bob are generic people in the group, by showing that they are the same age, I have shown that any two people in the group are the same age, therefore everyone in the group of k+1 is the same age, or s(k+1) = 1.

By induction, everyone in the world is the same age.

[Kit-]

Re: The final question on my alg2 exam last year.

Inductive hypothesis fails for k=2, since there's no overlap.

[blve23]

Re: The final question on my alg2 exam last year.

true but like i said I DIDNT DO ****!