Could someone prove this

I think a proof by contradiction should work.

First, let's let k = -b, with b > 0, for simplicity.

Now, given x < y, we know that 0 < y - x.

Then, assume that -bx < -by.

It follows:

-bx - (-by) < 0
-b(x-y) <0

Since b is non-zero, we may divide it out (I suppose this depends on the fact that bx < by implies x < y for b > 0, so I'm assuming this has already been proved).

Then,

-(x-y) = y - x < 0

But, our original statement is that y - x > 0, so we have a contradiction, which means that -bx > -by.

(I think this might also require the fact that we know -bx = -by is true only when x = y, which is again a violation of x < y, otherwise I don't think this proof alone rules out that possibility).

 

You don't need an argument by contradiction if you accept the "rule of signs" for multiplying numbers:

+ times + is +
+ times - is -
- times + is -
- times - is +

x < y
x-y < 0
If k < 0 then k(x-y) > 0 (minus times minus = plus)
kx - ky > 0
kx > ky

 

And where do you pull your "rule of signs" from?
It is a higher-order statement that the one to be proven.

 

I learned it in elementary school, about 50 years ago.

I assumed this forum it was intended to help people, not to show off ones own knowledge and/or have p*ssing competitions. Apologies if I was wrong about that.

If you want to post a proof starting from Peano's axioms and a definition of the real numbers using Dedekind cuts, feel free. But I doubt if the OP would benefit much from reading it.

 

This formum is not about getting huffy is someone objects to what you have done. The real problem is that the original poster did not show any work and so we have no idea what "basis" he is starting from- and so neither you nor I have any idea what the OP would benefit from! Certainly, if you are allowed to use the fract that "positive times positive is positive" the "proof" is trivial. That is why I would suspect that that is not what the OP wanted.

 

The rule of signs follows directly from the axioms for an ordered field.

Axiom 1: if a > b, then for any c, a+c > b+c

As a special case, take b = 0 and c = -a: then Axiom 1 becomes

If a > 0 then -a < 0
Similarly, if a < 0 then -a > 0

Axiom 2: if a > 0 and b > 0 then ab > 0

So to prove one case of the rule of signs:

if a < 0 and b > 0, then -a > 0 (axiom 1)
(-a)(b) > 0 (axiom 2)
-(ab) > 0 (Field axioms of arithmetic)
ab < 0 (axiom 1)

I don't understand Arildno's comment that the rule of signs is "a higher order statement than the one to be proven".

 

True. If someone were to ask me how to prove "if x< y then for any k< 0, ky< kx", I would probably start with the axiom: "if a> 0 and b> 0 then ab> 0", or, equivalently "if a< b and c> 0 then ac> bc". Of course, we still don't know what rhule009 has to work with!