Question: what are the conditions on k for ka + b/k = a+b ?

[Steven G]

I was thinking the other day how a*b = (ka)*(b/k) for any non zero k. Then I was thinking what are the conditions on k for ka + b/k = a+b.

Let's see who can do this without using algebra, ie just by thinking about it. Of course k=1 is one of the solutions.

 

[Deleted member 4993]

I was thinking the other day how a*b = (ka)*(b/k) for any non zero k. Then I was thinking what are the conditions on k for ka + b/k = a+b.

Let's see who can do this without using algebra, ie just by thinking about it. Of course k=1 is one of the solutions.

I can think in "algebra". Are we restricted to real domain?

 

[Dr.Peterson]

I was thinking the other day how a*b = (ka)*(b/k) for any non zero k. Then I was thinking what are the conditions on k for ka + b/k = a+b.

Let's see who can do this without using algebra, ie just by thinking about it. Of course k=1 is one of the solutions.

I'm not sure of the conditions. In the first equation, it's true for all a and b, for any nonzero k. Is the second also supposed to be true for all a and b, or just for some given a and b? I think it's the latter, but I have an answer for either question.

 

[Steven G]

I can think in "algebra". Are we restricted to real domain?

Real domain (and a is not 0)

 

[Steven G]

I'm not sure of the conditions. In the first equation, it's true for all a and b, for any nonzero k. Is the second also supposed to be true for all a and b, or just for some given a and b? I think it's the latter, but I have an answer for either question.

For a given a and b (a is not zero)

 

[Dr.Peterson]

For a given a and b (a is not zero)

Then the answer is that k can be either 1 or b/a.

Algebra easily proves that these are the only solutions; "inspection" can reveal both (since b/a changes a to b and b to a), but I'm not sure you can be certain that there are no others.

 

[Steven G]

Then the answer is that k can be either 1 or b/a.

Algebra easily proves that these are the only solutions; "inspection" can reveal both (since b/a changes a to b and b to a), but I'm not sure you can be certain that there are no others.

You can see that you will get a quadratic equation in k, so no more than 2 solutions. Isn't that enough?

 

[Dr.Peterson]

You can see that you will get a quadratic equation in k, so no more than 2 solutions. Isn't that enough?

You missed my point. You had said, "Let's see who can do this without using algebra, ie just by thinking about it." I said, "Algebra easily proves that these are the only solutions; "inspection" can reveal both (since b/a changes a to b and b to a), but I'm not sure [by inspection] you can be certain that there are no others."

It's often easy to just see a solution, but you need more (such as algebra) to be sure you've found all solutions.