Order axioms and x/y > 1 (For all x, y, z ∈ ℝ, if x<y and 0<z, then xz<yz)

[Four Muffins]

Hello.

I was working on a problem and made a mistake, and I'm trying to understand exactly what it was. I didn't realise when I was writing the solution that
x/y > 1 ---> x > y
doesn't work when x and y are negative.

Have I not followed the axiom
(3d) For all x, y, z ∈ ℝ, if x<y and 0<z, then xz<yz
from here, and that by using x and y with a -1 factor, I haven't flipped the inequality sign? Or is it something else, like a general math thing that axioms aren't needed for, and I just have to notice it doesn't work and qualify the statement with 'for x, y ∈ ℝ+'? Or is it just not valid to multiply both sides of the inequality by y?

 

[Dr.Peterson]

Hello.

I was working on a problem and made a mistake, and I'm trying to understand exactly what it was. I didn't realise when I was writing the solution that
x/y > 1 ---> x > y
doesn't work when x and y are negative.

Have I not followed the axiom
(3d) For all x, y, z ∈ ℝ, if x<y and 0<z, then xz<yz
from here, and that by using x and y with a -1 factor, I haven't flipped the inequality sign? Or is it something else, like a general math thing that axioms aren't needed for, and I just have to notice it doesn't work and qualify the statement with 'for x, y ∈ ℝ+'? Or is it just not valid to multiply both sides of the inequality by y?

Please show us the problem you are working on, which is not clearly stated here; and tell us why you think you are wrong.

When you say "Are x and y positive", is that a question, or are you trying to say "Suppose x and y are positive"?

Is it given that 2x - 2y = 1 and also that x/y > 1?

What is your goal? What was your conclusion that you think is wrong?

If you know that x and y are positive, and that x/y > 1, then you can multiply by y to conclude that x > y, because the axiom says you can. But if y can be negative, then you can't.

 

[Four Muffins]

I beg your pardon, I should have been clearer. That is the question, here's the source.

Inequality and absolute value questions from my collection

Guys I didn't forget your request, just was collecting good questions to post. So here are some inequality and absolute value questions from my collection. Not every problem below is hard, but there are a ...

gmatclub.com

And the solution is here: https://gmatclub.com/forum/inequali...ions-from-my-collection-86939-20.html#p653709


I was wondering if there was a particular rule or something that I violated, since using simple algebraic manipulation on x/y > 1 led me astray, with me not noticing that multiplying both sides of x/y > 1 by y doesn't work if x and y are both negative.

(Presumably 'Answer: C.' in the solution refers to an answer sheet I haven't found.)

Edit: missed words

 

[Otis]

if x and y are both negative

Hi Four Muffins. If x and y are both negative, then they each contain a factor of -1, which cancel. In other words, you end up with |x|/|y| which is the same as x and y both positive. Yet, the case where x and y are both negative doesn't work in this exercise because y<x yields |y|>|x| after the negations cancel in the ratio x/y.

Your example: x = -1 and y = -3/2

\(\displaystyle \frac{x}{y} = \frac{\cancel{(\text{-}1)}(1)}{\cancel{(\text{-}1)}(\frac{3}{2})} = \frac{2}{3}\)

That's the same case as x = 1 and y = 3/2, which is not a valid (x,y) pair because we know that y must be less than x to make x/y greater than 1.

---

We can confirm graphically that x and y must both be positive, whenever both Statement1 and Statement2 are true.

Statement1 tells us that y is the same as x – ½

Substitute that expression for y in Statement2:

x/(x – 1/2) > 1

Here's a graph of each side of that inequality. NOTE that the vertical axis is not called the y-axis in this plot; the vertical axis measures the value of ratio x/y, not the value of y.

The left-hand side is plotted in green and the right-hand side in yellow. We see that the green curve shows x/y greater than 1 (i.e., the green curve appears above the yellow line) only when x is greater than ½ (line x=½ shown in red). That means y must be positive, also, because y is the result of subtracting ½ from a number that's bigger than ½.



Alternatively, by taking cases, we can solve the inequality x/(x–½)>1 algebraically and find that it's true for all x>½.
[imath]\;[/imath]

 

[Dr.Peterson]

I beg your pardon, I should have been clearer. That is the question, here's the source.

Inequality and absolute value questions from my collection

Guys I didn't forget your request, just was collecting good questions to post. So here are some inequality and absolute value questions from my collection. Not every problem below is hard, but there are a ...

gmatclub.com

View attachment 34695

And the solution is here: https://gmatclub.com/forum/inequali...ions-from-my-collection-86939-20.html#p653709
View attachment 34696

I was wondering if there was a particular rule or something that I violated, since using simple algebraic manipulation on x/y > 1 led me astray, with me not noticing that multiplying both sides of x/y > 1 by y doesn't work if x and y are both negative.

(Presumably 'Answer: C.' in the solution refers to an answer sheet I haven't found.)

Edit: missed words

Apparently, the choices given are something like:

A: (1) alone is sufficient to answer​

B: (2) alone is sufficient to answer​

C: (1) and (2) together are sufficient to answer​

D: There is not enough information to know​

Omitting this was a major error in presenting the problem. I would not trust this person.

As for your work, you are correct that [imath]\frac{x}{y}>0[/imath] does not imply that [imath]x>y[/imath], so that is a mistake.

One alternative method to theirs would be to take two cases, according to the sign of y, allowing you to do your multiplication and know the result.

 

[Four Muffins]

That makes a lot of sense, thanks very much.

 

[Steven G]

If x-y>0 (as in x-y=1/2), then surely x>y.
The proof is trivial. In x-y>0, just add y to both sides to get x>y.

Now if x/y>1 we have two cases to consider.
Case 1: x and y are both positive.
If x/y >1, multiplying both sides by y, we get x>y.

Case 2: x and y are both negative.
In x/y>1, multiply both sides by y and get x<y
OR multiply x/y by -1/-1 to get -x/-y>1. Now multiply both sides by -y and get -x>-y. Multiplying both sides by -1 yields that x<y