MHB Calculate w/ Defective Calculator: Multiply w/ Add, Subtract, & Reciprocal

[Dethrone]

You have a malfunctioning calculator that cannot perform multiplication. However, it can add, subtract, and compute the reciprocal $\frac{1}{x}$ of any number $x$. Can you nevertheless use this defective calculator to multiply numbers?

Spoiler

Source: 3rd International Young Mathematicians' Convention

 

[I like Serena]

Hey Rido! ;)

What? My calculator isn't malfunctioning. It multiplies just fine. (Dull)

Oh well... grumbling... (Sweating)
I'll just do:

Spoiler

$$xy=\frac{x+y}{\frac 1x + \frac 1y}$$ (Whew)
(I recognize the form from parallel resistors. (Smirk))

 

[Dethrone]

Hey Rido! ;)

What? My calculator isn't malfunctioning. It multiplies just fine. (Dull)

Oh well... grumbling... (Sweating)
I'll just do:

Spoiler

$$xy=\frac{x+y}{\frac 1x + \frac 1y}$$ (Whew)
(I recognize the form from parallel resistors. (Smirk))

Very interesting! Thank you for participating. :D However, one point I am not following, and it might be because I don't own a malfunctioning calculator.

You have $x$ and $y$, and so you are able to computed $x+y$ (1) in the numerator. You were also able to take the reciprocal of $\frac{1}{x}$ and $\frac{1}{y}$, and add them to get $\frac{1}{x}+\frac{1}{y}$ (2). But, how were you able to combine step (1) and (2) without being able to multiply? (Wondering) Can you use your defective calculator to confirm?

 

[I like Serena]

Very interesting! Thank you for participating. :D However, one point I am not following, and it might be because I don't own a malfunctioning calculator.

You have $x$ and $y$, and so you are able to computed $x+y$ (1) in the numerator. You were also able to take the reciprocal of $\frac{1}{x}$ and $\frac{1}{y}$, and add them to get $\frac{1}{x}+\frac{1}{y}$ (2). But, how were you able to combine step (1) and (2) without being able to multiply? (Wondering) Can you use your defective calculator to confirm?

Oh darn. I guess I'll have to buy one of those defective calculators just to figure it out. (Crying)

 

[chisigma]

You have a malfunctioning calculator that cannot perform multiplication. However, it can add, subtract, and compute the reciprocal $\frac{1}{x}$ of any number $x$. Can you nevertheless use this defective calculator to multiply numbers?

[sp]Is...

$\displaystyle x\ y = \frac{x\ y}{x + y}\ (x + y) = \frac{x+y}{\frac{1}{x} + \frac{1}{y}}$

[/sp]

Kind regards

$\chi$ $\sigma$

 

[Nono713]

Spoiler

There is a bug (division by zero) in the solutions above if $x = -y$. If this is the case then the alternative version:
$$xy = \frac{xy}{x - y} (x - y) = \frac{x - y}{\frac{1}{y} - \frac{1}{x}}$$
should be used. At least one of the two versions will work except for $x = y = 0$ in which case $xy = 0$ and no calculator is required.

 

[Opalg]

You have a malfunctioning calculator that cannot perform multiplication. However, it can add, subtract, and compute the reciprocal $\frac{1}{x}$ of any number $x$. Can you nevertheless use this defective calculator to multiply numbers?

[sp]
Step 1.From $x$ you can get $\frac12x$. In fact, $\dfrac1{\frac1x + \frac1x} = \frac12x.$

Step 2. From $x$ you can get $x^2$. In fact, $\dfrac1{x-0.5} - \dfrac1{x+0.5} = \dfrac1{x^2 - 0.25}.$ The reciprocal of that is $x^2 - 0.25$. Add $0.25$ to get $x^2$.

Step 3. From $x$ and $y$ you can get $xy$. In fact, from $x$ and $y$ you can get $x+y$ (by addition), then $(x+y)^2 = x^2 + 2xy + y^2$ (by Step 2). Now subtract $x^2 + y^2$ (which again you can get by Step 2 together with addition) to get $2xy$. Finally, use Step 1 to get $xy.$[/sp]

 

[Dethrone]

Thanks for participating, Bacterius and $\chi$ $\sigma$, but I am pretty sure you are making an multiplication step in order amalgamate $ (x - y)$ and $\frac{1}{\frac{1}{y} - \frac{1}{x}}$.

@Opalg, excellent solution! (Cool) Thanks for participating.

 

[Dethrone]

Solution of other:

Spoiler

Solution:
To multiply $x$ by a negative number $-n^2$:
$$-n^2x=n+\frac{1}{-\frac{1}{n}+\frac{1}{n+\frac{1}{x}}}$$

To multiply $x$ by a positive number $n^2$:

$$n^2x=n+\frac{1}{-\frac{n+1}{n}+\frac{1}{1+\frac{1}{-(n+1)+\frac{1}{x}}}}$$

 

[Opalg]

Solution of other:

Spoiler

Solution:
To multiply $x$ by a negative number $-n^2$:
$$-n^2x=n+\frac{1}{-\frac{1}{n}+\frac{1}{n+\frac{1}{x}}}$$

To multiply $x$ by a positive number $n^2$:

$$n^2x=n+\frac{1}{-\frac{n+1}{n}+\frac{1}{1+\frac{1}{-(n+1)+\frac{1}{x}}}}$$

I'm dubious about "other"'s solution. If you want to find $xy$ (with $y$ positive) then this solution says that you need to express $y$ in the form $n^2$. But we are not told whether the calculator has a square root function (or if it has, whether that is also malfunctioning). So how do you find $n$?

 

[I like Serena]

I checked Opalg's solution on my new defective calculator, just to see how it looks: (Thinking)

Spoiler

$$
xy=\frac 1{\frac 1{\frac 1{\frac 1{x+y-\frac 12}-\frac 1{x+y+\frac 12}}
- \frac 1{\frac 1{x-\frac 12}-\frac 1{x+\frac 12}}
- \frac 1{\frac 1{y-\frac 12}-\frac 1{y+\frac 12}}
- \frac 14} + \frac 1{\frac 1{\frac 1{x+y-\frac 12}-\frac 1{x+y+\frac 12}}
- \frac 1{\frac 1{x-\frac 12}-\frac 1{x+\frac 12}}
- \frac 1{\frac 1{y-\frac 12}-\frac 1{y+\frac 12}}
- \frac 14}}
$$

So even on my defective calculator, I can still easily calculate a product. (Happy)

 

[Dethrone]

Hi Opalg, you are absolutely correct about that point. The solution used by the author makes use of the fact that the calculator has a functioning square root button. The author also presented another solution, but that one is identical to the one you posted.

I thought this might be a cool trick to show to friends...or in case of an emergency during a test when the calculator really breaks down...hehe :D

Spoiler

Although the possibility of just the multiplication button not working is just not likely. :p