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[ed boy]

we have:

4b^2 = a^2

so a² is a multiple of four.
We also have:

But doesn't the original proof rely on if N² is a multiple of x, N must also be?

Yes.  What's the problem?

so if a² is a multiple of four, a must also be a multiple of four. Yet:

a = 2c isn't a multiple of four.  a = 2.

[Vector]

so if a² is a multiple of four, a must also be a multiple of four.

...

[Pillow_Killer]

so if a² is a multiple of four, a must also be a multiple of four.

[ed boy]

You supported that part:

But doesn't the original proof rely on if N² is a multiple of x, N must also be?

Yes.  What's the problem?

Furthermore, the argument for the square root of two relies on it as well.

[Vector]

You supported that part:

But doesn't the original proof rely on if N² is a multiple of x, N must also be?

Yes.  What's the problem?

Furthermore, the argument for the square root of two relies on it as well.

Jesus Christ.

Yes, you're right, I spoke badly.

But in more seriousness, sit down like all the rest of us had to, and write out the proof, and figure out why it works.  It isn't that hard.

[ILikePie]

Forgive the interruption. I've been given a geometry problem that I can't seem to solve. I need to prove that if you connect the middles of a quadrilateral segments you get a rectangle. I also know that the diagonals of said quadrilateral are perpendicular to each other (90 degrees).
Any ideas?

[Vector]

Doesn't perpendicularity of diagonals imply that you have a kite?

[ILikePie]

Indeed it does, though, it doesn't help much with the solution.

e, I'll write it down. I have a points ABCD which create a kite. Points E, F,G, and H, divide AB, BC, CD and DA in half respectively. I need to prove that EFGH is a rectangle.

[Jim Groovester]

Indeed it does, though, it doesn't help much with the solution.

WHAAAAAT.

Yes it does.

Anyways, if you're stuck, just prove the connecting segments are parallel to the diagonals by using similar triangles.

[Vector]

Indeed it does, though, it doesn't help much with the solution.

Sure it does... it solves the entire problem for us, because we end up with a whole bunch of congruent right triangles.

[Virex]

so if a² is a multiple of four, a must also be a multiple of four. Yet:

a² = m*4, m is an integer
a = (m*4)^1/2
a = 2*m^1/2
m^1/2 is not necessarily an integer rational, nor is 2*m^1/2 necessarily a multiple of 4 (consider the case for m = 2 for example)

[ILikePie]

Indeed it does, though, it doesn't help much with the solution.

Sure it does... it solves the entire problem for us, because we end up with a whole bunch of congruent right triangles.

I still don't see it, and it's late. I'll ask some people in class tomorrow, I'm sure someone got it right. Thanks for the help you two.

[KaminaSquirtle]

Woohoo, I figured out my problem!  I'm about 1/2 to 1/3 done writing the proof out.  It feels so good to have finally solved it!

[Vector]

Woohoo, I figured out my problem!  I'm about 1/2 to 1/3 done writing the proof out.  It feels so good to have finally solved it!

Managed to get someone to explain my problem to me... turns out that the main reason why I was being so dumb about it was a serious notation issue.  Never would have thought to write it out that way

Bah-bleeding-humbug.

[RedWarrior0]

Forgive the interruption. I've been given a geometry problem that I can't seem to solve. I need to prove that if you connect the middles of a quadrilateral segments you get a rectangle. I also know that the diagonals of said quadrilateral are perpendicular to each other (90 degrees).
Any ideas?

I remember these. First of all, you'll always get a parallelogram when you connect the midpoints of adjacent sides of a quadrilateral.
Given Quadrilateral ABCD, with midpoint of AB at F and midpoint of BC at G. Draw diagonal AC and segment FG. Since FB:AB = GB:CB = 1/2, FG:AC = 1/2. Do the same with the other side and diagonal, and you will find that you always get a parallelogram with sides parallel to the diagonals of the quadrilateral, and you can take it from there. Hopefully.