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## Homework Statement

show that if x and y are real numbers such that x<y, then for any real number k<0, kx>ky

## Homework Equations

## The Attempt at a Solution

note- this is not a homework. im just teaching myself some algebra, so please help

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- Thread starter rhule009
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- #1

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show that if x and y are real numbers such that x<y, then for any real number k<0, kx>ky

note- this is not a homework. im just teaching myself some algebra, so please help

- #2

Mute

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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).

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

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

Then,

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).

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- #3

AlephZero

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+ 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

- #4

arildno

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It is a higher-order statement that the one to be proven.

- #5

AlephZero

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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.

- #6

HallsofIvy

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- #7

AlephZero

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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".

- #8

HallsofIvy

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