2.3.4. Prove the identity \((a+b)^2=a^2+2ab+b^2\) for all natural numbers \(a,b\).
- \(\bullet\) Show that \((a+b)^2=a^2+2ab+b^2\) for all natural numbers \(a,b\).
- \(\Vdash\) \((a+b)^2\)
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\(=\) { Definition of exponentiation: \(m^0:=1\); \(m^{n\pp}:=m^n\times m\);
\(\hspace{1cm} m^2=m^{1\pp}=m^1\times m=m^0\times m\times m=1\times m\times m=m\times m\) }- \((a+b)\times (a+b)\)
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\(=\) { Distributive law: \((b+c)a=ba+ca\) }
- \(a(a+b)+b(a+b)\)
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\(=\) { Distributive law: \(a(b+c)=ab+ac\) }
- \(aa+ab+ba+bb\)
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\(=\) { Commutative law: \(nm=mn\) }
- \(aa+ab+ab+bb\)
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\(=\) { Definition of multiplication: \((n\pp)\times m:=n\times m+m\);
\(\hspace{1cm} 2m=(1\pp)m=1m+m=m+m\) }- \(aa+2ab+bb\)
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\(=\) { Definition of exponentiation: \(m^0:=1\); \(m^{n\pp}:=m^n\times m\);
\(\hspace{1cm} a^2=a^{1\pp}=a^1a=a^0aa=1aa=aa\) }- \(a^2+2ab+b^2\)
- \(\square\)