Are there any numbers for which this true?

>he doesnt solve for d

what is even the point

Holy shit. Don't mind me I'm just gonna shoot myself real quick

Works whenever

[math] C = - A\, (\frac{D}{B})^2 [/math]

For example

[math] A = 3, B = 2, D = 4 [/math]

[math] C = -3 (\frac{4}{2})^2 = -12 [/math]

So

[math] \dfrac { 3 } { 2 } + \dfrac { -12 } { 4 } = 1.5 + (-3) = -1.5 [/math]

and indeed

[math] \dfrac { 3+(-12) } { 2+4 } = \dfrac { -9 } { 6 }= -1.5 [/math]

It's true for certain 4x4 matrices.

Show me.

en.wikipedia.org/wiki/Farey_sequence

if A=B=C=D, it works.

No sir

Great work man, just wondering how did you get this? I tried back tracking and got this
A/B + C/D = (A+C)/(B+D)
(A/B + C/D)(B+D) = A + C
A + AD/B + CB/D + C = A + C
AD/B + CB/D = 0
(AD/B + CB/D)(BD) = 0(BD)
(AD)^2 + (CB)^2 = 0
(CB)^2 = -(AD)^2
(CB)(CB)(1/CB^2) = -1(AD)(AD)(1/CB^2)
--------------------------------------------------
AD/B = -(CB/D)
AD = -(CB^2/D)
------------------------------------------------
C = -1(-1)(CB^2/D)(-1)(CB^2/D)(1/CB^2)
C = -1(CB^2/D)(1/D)
C = -1(C/D)(B^2/D)(1/D)
C = -1(C/D)(B^2/D^2)
------------------------------------------------
A/B + C/D = (A+C)/(B+D)
C/D = (A+C)/(B+D) - A/B

and essentially from here I am stuck

Sorry for the lack of latex on mobile

he said numbers

Actually, no

If A=0, C=0, B and D can be any other number and it works. It's the only way.

Wrong.

no, man.

wtf

m/n +/- r/s = (ms+/-nr)/ns

...

a/b +c/d = (a+c)/(b+d).
(ad + bc)/bd = (a+c)/(b+d).
bd(a+c) = (b+d)(ad+bc).
ad2 + b2c = 0.

so, any numbers that the following three conditions are true:
1)b,d != 0.
2)b != -d.
3)ad2 + b2c = 0.

gtfo pleb

>Choose any b c and d, plug in to get a
Let b = c = 0, good luck pleb

L0L fgt pls

>implying zero is not a number
back to school fgt pls

>b != -d
wat

for [math]BD \neq 0[/math]
[math] \displaystyle A=- \left ( \frac{B}{D} \right )^2 \times C[/math]

One line in Mathematica (sorry if that answer disappoints you)

No problem!

A=1 C=-1

B= 2 D=2

Try some of these:

{a -> 0, b -> -3965, c -> 0, d -> -1}, {a -> -978, b -> -1, c -> 978,
d -> -1}, {a -> 0, b -> 4929, c -> 0, d -> 26}, {a -> -3434, b -> -1,
c -> 3434, d -> -1}, {a -> 0, b -> -712, c -> 0,
d -> -93}, {a -> 3611, b -> -94, c -> -3611, d -> -94}, {a -> 7542,
b -> 1, c -> -7542, d -> 1}, {a -> 0, b -> -548, c -> 0,
d -> 600}, {a -> 6750, b -> -13, c -> -6750, d -> -13}, {a -> 0,
b -> -1458, c -> 0, d -> -32}, {a -> -907, b -> 11, c -> 907,
d -> 11}, {a -> 1003, b -> 92, c -> -1003, d -> 92}, {a -> 0,
b -> -3491, c -> 0, d -> -1}, {a -> -7717, b -> 1, c -> 7717,
d -> 1}, {a -> 8454, b -> -58, c -> -8454, d -> -58}, {a -> 0,
b -> -4297, c -> 0, d -> -1}, {a -> -5254, b -> -1, c -> 5254,
d -> -1}, {a -> -9950, b -> -85, c -> 9950, d -> -85}, {a -> 0,
b -> 2836, c -> 0, d -> -2880}, {a -> 0, b -> -6348, c -> 0,
d -> 6360}, {a -> 0, b -> 8412, c -> 0, d -> -8410}, {a -> -4152,
b -> -1, c -> 4152, d -> -1}, {a -> -8993, b -> -9, c -> 8993,
d -> -9}, {a -> 0, b -> -2649, c -> 0, d -> -23}, {a -> 0,
b -> -1606, c -> 0, d -> -45}, {a -> -6661, b -> 95, c -> 6661,
d -> 95}, {a -> 0, b -> 8699, c -> 0, d -> 1}, {a -> -3242, b -> 49,
c -> 3242, d -> 49}, {a -> 0, b -> -4115, c -> 0, d -> 93}, {a -> 0,
b -> -2408, c -> 0, d -> 91}, {a -> 802, b -> -84, c -> -802,
d -> -84}, {a -> -978, b -> -47, c -> 978, d -> -47}, {a -> 0,
b -> 6102, c -> 0, d -> -6063}, {a -> -2670, b -> -10, c -> 2670,
d -> -10}, {a -> 523, b -> -91, c -> -523, d -> -91}, {a -> 945,
b -> 2, c -> -945, d -> 2}, {a -> 7899, b -> -1, c -> -7899,
d -> -1}, {a -> 0, b -> 6400, c -> 0, d -> -6422}, {a -> -1906,
b -> 35, c -> 1906, d -> 35}, {a -> 0, b -> -7616, c -> 0,
d -> -1}, {a -> 0, b -> 2597, c -> 0, d -> -2506}, {a -> -2745,
b -> -1, c -> 2745, d -> -1}, {a -> 0, b -> -6047, c -> 0,
d -> -1}, {a -> 2883, b -> 1, c -> -2883, d -> 1}, {a -> 0,
b -> -5844, c -> 0, d -> -1}, {a -> -8394, b -> -78, c -> 8394,
d -> -78}

>(AD)^2 + (CB)^2 = 0
is wrong. Only D and B are squared.

If B = D = 2
Then as long as A = -C it works

1

Are you guys retarded?

>1+1=1

Relly meks yu thunk

The link between the four number is : cb2 + ad2 = 0. The subset of |R that consists of the set of solutions cannot be represented visually.
However, we can "cut" into this representation by setting given values, for, say, a.

For a=1, here is the set (meaning b on the x axis-,c on the y axis, d on the z axis) of numbers that are solutions of what user is asking.

We can do more cuts, for different values of a. We even could study the "shape" in terms of values of a I guess.

note that for the mediant to even be well defined, we may assume the denominators are > 0 anyway, see en.wikipedia.org/wiki/Mediant_(mathematics)

therefore we have (b,y>0)

a/b + x/y = (a+x)/(b+y)
iff a/b^2 = -x/y^2

which is really the same as btw @ the user from literally kys m8. it's 2 lines of highschool-level calculation

same goes for this faggot >all those 0+0=0 entries
lmfao pic related

...

A=1
C=-1

C and D can be whatever but hey need to be equal to each other

i'm bit drunk so Ican't be 100% sure though

oops meant B and D need to be equal

Wow. Thank you, Master, to reformulate what many others have. Thanks so much.
Now. Can you help us go a little bit onward, or are you happy with this result?
>a/b^2 = -x/y2
Or maybe it is too difficult for you and you don't know how to address the problem?

Hasn't this guy given the clues?Cheers, brainlet. Next time you wanna give lessons to others, think first. Speak then.

sorry I was wrong I'm too drunk

CB^2+AD^2=0
B=/=0
D=/=0
B+D=/=0

Any pythagorean triple works.

There are 4 numbers involved, dude.
Please state your mathematical sentence more clearly. Or shut the fuck up.

except that he didn't said it was a mediant, he just asked the solution to the equation, therefore, my answer holds

what about pythagorean quadruples?

a/b + c/d = a+c/b+d

a/b = c/d + a+c/b+d

a = b* (c/d + a+c/b+d)

a = b*c/d + b*c/b+d +[ b*a/b+d ]

a >= b*c/d + b*c/b+d

x = b*c/d + b*c/b+d

a = x + b*x/b+d +[ a-x*b/b+d ]

whaddabout'em fgt pls

x = 1*1/1 + 1*1/1+1

a = 1/1+1/2 + 1*1.5/2 + ?-1.5 / 2

? = 0, a = 0.75 (lol)
? = 0.75, a=1.5 (again?)
?= 1.5, a = 2.25
?=2.25, a = 2.25 + 0.375
?=2.625, a = 2.625 + 0.5625

maybe if d >> b

not the pythag guy guy

if [math]\displaystyle \frac{A}{B}< \frac{C}{D}[/math] then
[math]\displaystyle \frac{A}{B}< \frac{A+C}{B+D}< \frac{C}{D}[/math]
prove me wrong

Prove yourself right

A = B = -1
C = 1000
D = 10

A/B = 1
(A+C)/(B+D) = 111
C/D = 100

really made me think

only if AD^2 + B^2C = 0

[math]
A = t r^2\\
B = r\\
C = -t s^2\\
D = s
[/math]
with the constraint that [math]r \neq 0 \wedge s \neq 0 \wedge r+s \neq 0 [/math].

No, but I can use your mom, who is a zero.

Just invent some new numbers and call them complex or imaginary.

(-1/-1)+(-1/-1)= (1)+(1) = 2

(-1+ -1)/(-1+ -1)= (-2)/(-2) = 1

2 doesn't equal 1
Does this mean I can do maths?

Troll?
(1/1)+(1/1)=(1)+(1)=2
(1+1)/(1+1)=(2)/(2)=1

It's rather easy to prove with A, B, C, D > 0.

A/B = X/(B+D) ; X = (AB + AD)/B = A + AD/B

C/D = Y/(B+D) ; Y = (CB + CD)/D = C + CB/D

So :
A + AD/B < A + C < CB/D + C

Which means :
AD/B < C A/B < C/D (true)
And
CB/D > A C/D > A/B (true)

So the inequality is true.

>bad dad bbc
what did you mean by this?

Oh boy... I kekked so hard

fckn awesome

...

however it is not necessarily true for all reals

CE

Hence the A, B, C, D > 0 conditions.

If negative numbers are in the equation, it's more complicated to do order relations.
However I'm fairly certain you could establish a stronger result by taking |A/B| < |C/D|.

>|A/B| < |C/D|.

CE it may be possible with relations between A, C and B, D, or A, D and B, C

After thinking about it, I suppose you just need a fraction in the normal form (ie B and D strictly positive).
I'm too lazy to prove it tho.