# Analog multipliers? Dividers?

#### Zeeus

Joined Apr 17, 2019
468
Please can someone assist in making this.....?

$a^2$ and $a\cdot b$ and $\frac {c}{d}$

lol..and not latex

Transistors ?
a^2 function
a. b function

and x/y function

Thanks

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#### Papabravo

Joined Feb 24, 2006
12,563
This is Homework Help, not homework done for you. Where is the demonstration of the effort you have expended so far on this problem?

#### Zeeus

Joined Apr 17, 2019
468
Okay : will learn it : but due in 39hrs

#### Papabravo

Joined Feb 24, 2006
12,563
AFAIK there is nothing in the problem about mixers and opamps. There is nothing for you to learn except basic algebra to attempt the proof, and this is an emergency on my part -- why exactly?

#### Papabravo

Joined Feb 24, 2006
12,563
Will somebody please explain how the proof that

$$\frac{a^2\;+\;b^2}{ab\;+\;1}\;=\;c^2$$

has any connection to circuits

Oh great -- a new way to show errors is to center the TeX input in a little rectangular box. The old was far better. I'm just sayin'
If it is not toooo awful much trouble could you fix the TeX rendering in the blogs as well?

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#### Zeeus

Joined Apr 17, 2019
468
Will somebody please explain how the proof that
$$\frac{a^2\;+\;b^2}(ab\+\;1}=c^2$$
has any connection to circuits

Oh great -- now TeX rendering is well and truly broken -- great job guys
If it is not toooo awful much trouble could you fix it in the blogs as well?
No no...The proof has nothing to do with circuits : just need the multiplier circuit to test the result

#### Papabravo

Joined Feb 24, 2006
12,563
I see. Thanks for the clarification. Testing the proposition with analog circuits will certainly be a challenge considering that the theorem as stated is true only for integers. It is not necessarily true in the domain of real numbers. Are you ignoring this or did you overlook that detail?

I also observe than none of the 16 Pythagorean Triples with c <= 100 satisfy the condition.

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#### Zeeus

Joined Apr 17, 2019
468
I see. Thanks for the clarification. Testing the proposition with analog circuits will certainly be a challenge considering that the theorem as stated is true only for integers. It is not necessarily true in the domain of real numbers. Are you ignoring this or did you overlook that detail?
Suppose we apply a sine wave as input 1 and 0v as input 2 : what's output if not a sine wave squared?

or 0v and 2v...output is 4v?

#### Jony130

Joined Feb 17, 2009
5,023
$\frac{a^2\;+\;b^2}(ab\+\;1}=c^2$
s any connection to circuits
Oh great -- now TeX rendering is well and truly broken -- great job guys
If it is not toooo awful much trouble could you fix it in the blogs as well?
TeX works just fine.

$\frac{a^2+ b^2}{ab+1}=c^2$

#### Papabravo

Joined Feb 24, 2006
12,563
TeX works just fine.

$\frac{a^2+ b^2}{ab+1}=c^2$
OK -- it does have a new way of showing errors, and it still doesn't work in the blogs. Also everything is now centered and noticeably smaller.

#### Papabravo

Joined Feb 24, 2006
12,563
Suppose we apply a sine wave as input 1 and 0v as input 2 : what's output if not a sine wave squared?

or 0v and 2v...output is 4v?
I'm not sure what you are getting at here. A sine wave varies continuously between a positive peak value and a negative peak value. At some point it must be zero. What happens when you divide something by zero? There is literally no connection between integers and sine waves.

BTW can you find any integers a and b for which the proposition is true?

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#### Zeeus

Joined Apr 17, 2019
468
BTW can you find any integers a and b for which the proposition in true?
No can not except the trivial cases with a or b = 0 or a = b then c = 1 and a = bc

The proof uses "infinite descent" :
Assuming there's a solution (a,b) with minimum a then from vieta jumping..........
Let's leave the math for 'high school' kids

Wanted to test something but pointless now : back to Electronics stuffs 120v, buck converters..etc

Actually google "vieta jumping" the same question is there...if the denominator is changed to ab-1 then yes possible to find solutions except trivial cases

and sorry late response : not great day

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#### Papabravo

Joined Feb 24, 2006
12,563
No can not except the trivial cases with a or b = 0 or a = b then c = 1 and a = bc

The proof uses "infinite descent" :
Assuming there's a solution (a,b) with minimum a then from vieta jumping..........
Let's leave the math for 'high school' kids

Wanted to test something but pointless now : back to Electronics stuffs 120v, buck converters..etc

Actually google "vieta jumping" the same question is there...if the denominator is changed to ab-1 then yes possible to find solutions except trivial cases

and sorry late response : not great day
Thanks for the link. I had not heard of the method before, especially in regards to hyperbolas in quadrant 1.

#### Zeeus

Joined Apr 17, 2019
468
Thanks for the link. I had not heard of the method before, especially in regards to hyperbolas in quadrant 1.
You're in high school? college?