No algebra permitted

[bonanova]

A HS Freshman class solved a problem that described a group of children with their pet dogs.

They were told the number H of heads and the number F of feet belonging to the group.

They were asked for the number C of children and the number D of dogs.

 

Every student except Joe cranked out the solution using algebra.

Joe nevertheless was the first one with the answer.

 

If you eliminated the use of math, how would you solve this problem?

[k-man]

I don't know if there is any way to completely eliminate the use of math, but we can avoid equations.

 

First, barring any severed legs or feet, number F should be even. Let's divide it by 2. The resulting number F/2 accounts for every child once and for every dog twice.

H accounts for every child once and every dog once, so F/2 - H will be the number of dogs D. Subtract D from H to get C.

[bonanova]

I don't know if there is any way to completely eliminate the use of math, but we can avoid equations.

 

... so F/2 - H will be the number of dogs D ....

 

That's an equation, in words. And symbols are algebra.

 

The student found what might be called a constructive solution.

[witzar]

That's an equation, in words. And symbols are algebra.

The problem is to find d and c given h=(c+d) and f=(2c+4d).

As k-man stated, it can can be solved as follows: d=0.5f-h and c=2h-0.5f.

Suppose a solution of the problem exists that is "without algebra, equations, etc.".

Such a solution would be effectively a magic black box: you drop h and f inside,

and the box spits out (0.5f-h) and (2h-0.5f).

The magic box works like this:

(a,b) --> (a/2-b),(2b-a/2)

How can we use it?

1. (0,b) --> (-b,2b)

2. (a,0) --> (a/2,-a/2)

So we can use the box to multiply and divide an arbitrary number by two.

We can also use the box to do an aritrary subtraction (a-b):

first throw (0, a) in to get 2a, then throw (2a, b) in to get (a-b).

Of course we can do an arbitrarary addition too: a+b = a-(-b), so we can use (1.)

to get (-b) and reduce the problem to subtraction.

Having addition we do have a multiplication, since a*n = a+a+...+a (n times.).

An so on.

The question follows:

Could such magix box that allows arbitrary addition and subtraction

"without albebra, equations, etc." exist?

Or let me state it this way. Suppose Joe is your pal eager to solve

any children-and-dogs problem on your request. In such case you could

solve arbitrary arithmetical problems using only finite requests to Joe.

If Joe solves each such request "withoul algebra, equations, etc.",

then so do you. But is it possible?

Edited September 26, 2014 by witzar

[plasmid]

Joe was faster than the mathematicians because he thought that he was in art class instead of math class.

[bonanova]

 

That's an equation, in words. And symbols are algebra.

 

The problem is to find d and c given h=(c+d) and f=(2c+4d).

As k-man stated, it can can be solved as follows: d=0.5f-h and c=2h-0.5f.

Suppose a solution of the problem exists that is "without algebra, equations, etc.".

Such a solution would be effectively a magic black box: you drop h and f inside,

and the box spits out (0.5f-h) and (2h-0.5f).

The magic box works like this:

(a,b) --> (a/2-b),(2b-a/2)

How can we use it?

1. (0,b) --> (-b,2b)

2. (a,0) --> (a/2,-a/2)

So we can use the box to multiply and divide an arbitrary number by two.

We can also use the box to do an aritrary subtraction (a-b):

first throw (0, a) in to get 2a, then throw (2a, b) in to get (a-b).

Of course we can do an arbitrarary addition too: a+b = a-(-b), so we can use (1.)

to get (-b) and reduce the problem to subtraction.

Having addition we do have a multiplication, since a*n = a+a+...+a (n times.).

An so on.

The question follows:

Could such magix box that allows arbitrary addition and subtraction

"without albebra, equations, etc." exist?

Or let me state it this way. Suppose Joe is your pal eager to solve

any children-and-dogs problem on your request. In such case you could

solve arbitrary arithmetical problems using only finite requests to Joe.

If Joe solves each such request "withoul algebra, equations, etc.",

then so do you. But is it possible?

 

 

plasmid's method is what actually happened in the class.

It involves only counting and comparing.

 

But honorable mention to you for tenaciously following an out of the box approach.