Let * be a binary operation on set of rational number QQ defined as follows:a*b=a+b+2ab, where a,b in QQa) Prove that * is commutative, associate algebraic operation on QQ

FizeauV 2020-12-16 Answered

Let \(\times\) be a binary operation on set of rational number \(\displaystyle\mathbb{Q}\) defined as follows: \(a\cdot b=a+b+2ab\), where \(\displaystyle{a},{b}\in\mathbb{Q}\)
a) Prove that \(\times\) is commutative, associate algebraic operation on \(\displaystyle\mathbb{Q}\)

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Expert Answer

Mitchel Aguirre
Answered 2020-12-17 Author has 17664 answers

a) For commutative
Prove \(a\cdot b=b\cdot a\)
Now \(a\cdot b=a+b+2ab\)
and \(b\cdot a=b+a+2ba\)
\(b\cdot a=a+b=a+b+2ab\)

\(\displaystyle{\left\langle{a}+{b}={b}+{a}{\quad\text{and}\quad}{a}\cdot{b}={b}\cdot{a}\right\rangle}\)
Hence \(a\cdot b=b\cdot a\)
For associatvie
Prove\((a\cdot b)\cdot c=a\cdot(b\cdot c)\)
Now\((a\cdot b)\cdot c=(a+b+2ab)\cdot c\)
\(=a+b+2ab+c+2(a+b+2ab)c\)
\(=a+b+2ab+2ac+abc+4abc\)
and \(a\cdot (b\cdot c)=a\cdot (b+c+2bc)\)
\(=a+b+c+2bc+2a(b+c+2bc)\)
\(=a+b+c+2bc+2ab+2ac+4abc\)
\(=a+b+c+2ab+2ac+2bc+4abc\)
Hence \((a\cdot b)\cdot c=a\cdot(b\cdot c)\)

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