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    If I have ( x(AB)/y(D) ) > ( x(C)/y(D) ), can I simplify it, by removing the x and y, to get AB/D > C/D?

    Provided that is the correct, if I were to use the reciprocal of ^-1, I know each side would flip, but would the inequality sign reverse as well, to get D/AB < D/C?
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    (Original post by Aear)
    If I have ( x(AB)/y(D) ) > ( x(C)/y(D) ), can I simplify it, by removing the x and y, to get AB/D > C/D?

    Provided that is the correct, if I were to use the reciprocal of ^-1, I know each side would flip, but would the inequality sign reverse as well, to get D/AB < D/C?
    For the first question: What you are doing, by "removing the x and y", is multiplying both sides of the inequality by y/x. You can do that, but if y/x is negative (i.e. if x and y have different signs), then you must reverse the inequality sign. This is due to the general rule that when both sides of an inequality are multiplied or divided by a negative quantity, it is necessary to reverse the inequality sign.

    For the second question: When you reciprocate both sides of an inequality, it is indeed necessary to reverse the inequality sign.
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    (Original post by Aear)
    If I have ( x(AB)/y(D) ) > ( x(C)/y(D) ), can I simplify it, by removing the x and y, to get AB/D > C/D?

    Provided that is the correct, if I were to use the reciprocal of ^-1, I know each side would flip, but would the inequality sign reverse as well, to get D/AB < D/C?
    \dfrac{x\left(AB\right)}{y\left(  D\right)} &gt; \dfrac{x\left(C\right)}{y(D)} I'm sure you can't cancel? because isn't that the same as

    \dfrac{ABx}{Dy} &gt; \dfrac{Cx}{Dy} in which case you could just get rid of the denominator but you can't cancel like you suggested i think
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    (Original post by HapaxOromenon3)
    For the first question: What you are doing, by "removing the x and y", is multiplying both sides of the inequality by y/x. You can do that, but if y/x is negative (i.e. if x and y have different signs), then you must reverse the inequality sign. This is due to the general rule that when both sides of an inequality are multiplied or divided by a negative quantity, it is necessary to reverse the inequality sign.

    For the second question: When you reciprocate both sides of an inequality, it is indeed necessary to reverse the inequality sign.
    What I'm doing is Euclidean Geometry, where AB, C, and D are line lengths, while x and y are line multipliers, that is, only increasing their length. They won't be negative numbers. The proposition in question is trying to prove that (AB: D)>(C: D) where AB>C. I wanted to be sure I could cancel those multipliers. It also is trying to prove that (D:C)>(D:AB). I know you can write ratios as fractions to get the same answer (simply a difference in the actual writing, I believe), so to acquire this the reciprocal is the only way I can see of doing this. Playing with it in ration form is too confusing. If what you say is true, then it all works.
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    (Original post by Aear)
    What I'm doing is Euclidean Geometry, where AB, C, and D are line lengths, while x and y are line multipliers, that is, only increasing their length. They won't be negative numbers. The proposition in question is trying to prove that (AB: D)>(C: D) where AB>C. I wanted to be sure I could cancel those multipliers. It also is trying to prove that (D:C)>(D:AB). I know you can write ratios as fractions to get the same answer (simply a difference in the actual writing, I believe), so to acquire this the reciprocal is the only way I can see of doing this. Playing with it in ration form is too confusing. If what you say is true, then it all works.
    Yes, if everything is positive then your argument works.
 
 
 
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