To prove: [a, b] &gt; [c, d] if and only if abd2 − cdb2 ∈D+. Given information: Acoording to the definition of "greater than," &gt; is defined in Q by [a, b] &gt; [c, d] if and only if [a, b] − [c, d]∈Q+

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To prove: [a, b] &amp;gt; [c, d] if and only if abd2 − cdb2 ∈D+.  Given information:  Acoording to the definition of &quot;greater than,&quot; &amp;gt; is defined in Q by [a, b] &amp;gt; [c, d] if and only if [a, b] − [c, d]∈Q+

Arham Warner · Accepted Answer

Formula used: 1) Greatest than definition: Let D be an ordered integral domain with D+ as the set of positive elements. The relation greater than, denoted by &amp;gt;, is defined on elements x and y of D by x &amp;gt; y if and only if x − y ∈D+. 2) Property of greater that (&amp;gt;): If x &amp;gt; y and z &amp;gt; 0,then xz &amp;gt; yz Proof: Let [a,b] &amp;gt; [c,d] Then [a,b]−[c,d]∈ Q+ So,[a,b] + [−c,d]∈Q+ Therefore,[ad − bc,bd]∈Q+ This implies that(ad − bc)bd∈D+ So,adb2 − cdb2∈D+ Conversely, Let abd2 − cdb2∈D+ Therefore,(ad − bc)bd∈D+ So,[ad − bc,bd]∈Q+as bd≠q0 Then[a,b] + [−c,d]∈Q+ Implies that[a,b] − [c,d]∈Q+ Hence,[a,b] &amp;gt; [c,d] Therefore,[a,b] &amp;gt; [c,d]if and only if abd2 − cdb2∈D+

To prove: [a, b] > [c, d] text{if and only if} abd^2 - cdb^{2} in D^+. Given information: text{Acoording to the definition of "greater than,"} > text{is defined in Q by} [a, b] > [c, d] text{if and only if} [a, b] - [c, d] in Q^{+}

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