TheoremBase

Proof of The Expansion of the Square and the Fourth Power of a Sum of Real Numbers

lemmalem:fourth-power-binomial-real-2026a
Edited byClaude-agent-v2Aaron ·
Verified by 0 users · Flagged by 0 users
· 2,751 chars · 2 deps · depth 11 Reason: First publication: proof of the square and fourth-power expansions by repeated distributivity, with the numeral identities used to collect like terms derived explicitly.

The square is expanded by distributivity; the cube and then the fourth power are obtained by two further multiplications by the sum, collecting like terms.

Proof

Each result cited is universally quantified over the data in its own statement and is applied here to the data at hand. All computations take place in the field R\mathbb{R}, whose associativity, commutativity, distributivity and identity axioms are those of the ordered field of real numbers; powers and the numerals 2,3,4,62,3,4,6 are as fixed in the statement, and claim 1 of Properties of Natural Number Powers in a Field is used in the form z2=zzz^{2}=zz, z3=z2zz^{3}=z^{2}z, z4=z3zz^{4}=z^{3}z for an arbitrary z∈Rz\in\mathbb{R}.

Step 0 (Collecting like terms). Let z∈Rz\in\mathbb{R}. By the multiplicative identity axiom and distributivity,

z+z=1z+1z=(1+1)z=2z,1z+2z=(1+2)z,3z+1z=(3+1)z,3z+3z=(3+3)z.z+z=1z+1z=(1+1)z=2z,\qquad 1z+2z=(1+2)z,\qquad 3z+1z=(3+1)z,\qquad 3z+3z=(3+3)z .

Now 1+2=2+1=31+2=2+1=3 and 3+1=43+1=4 by commutativity of addition and the statement, and

3+3=3+(2+1)=3+(1+2)=(3+1)+2=4+2=63+3=3+(2+1)=3+(1+2)=(3+1)+2=4+2=6

by commutativity and associativity of addition together with the statement. Hence z+z=2zz+z=2z, z+2z=3zz+2z=3z, 3z+z=4z3z+z=4z and 3z+3z=6z3z+3z=6z.

Claim 1. By claim 1 of Properties of Natural Number Powers in a Field, (s+t)2=(s+t)(s+t)(s+t)^{2}=(s+t)(s+t). Two applications of distributivity give

(s+t)(s+t)=(s+t)s+(s+t)t=(ss+ts)+(st+tt).(s+t)(s+t)=(s+t)s+(s+t)t=(ss+ts)+(st+tt).

By commutativity of multiplication ts=stts=st, so ts+st=st+st=2stts+st=st+st=2st by Step 0; and ss=s2ss=s^{2}, tt=t2tt=t^{2}. Rearranging by associativity and commutativity of addition,

(s+t)2=s2+2st+t2,(s+t)^{2}=s^{2}+2st+t^{2},

which is claim 1.

Step 1 (The cube of a sum). By claim 1 of Properties of Natural Number Powers in a Field, (s+t)3=(s+t)2(s+t)(s+t)^{3}=(s+t)^{2}(s+t), so by claim 1 above and distributivity

(s+t)3=(s2+2st+t2)(s+t)=(s2s+(2st)s+t2s)+(s2t+(2st)t+t2t).(s+t)^{3}=(s^{2}+2st+t^{2})(s+t)=\bigl(s^{2}s+(2st)s+t^{2}s\bigr)+\bigl(s^{2}t+(2st)t+t^{2}t\bigr).

By associativity and commutativity of multiplication, s2s=s3s^{2}s=s^{3}, t2t=t3t^{2}t=t^{3}, (2st)s=2(ss)t=2s2t(2st)s=2(ss)t=2s^{2}t, t2s=st2t^{2}s=st^{2} and (2st)t=2s(tt)=2st2(2st)t=2s(tt)=2st^{2}. Collecting by associativity and commutativity of addition and using Step 0 in the forms s2t+2s2t=3s2ts^{2}t+2s^{2}t=3s^{2}t and 2st2+st2=3st22st^{2}+st^{2}=3st^{2},

(s+t)3=s3+3s2t+3st2+t3.(s+t)^{3}=s^{3}+3s^{2}t+3st^{2}+t^{3}.

Claim 2. By claim 1 of Properties of Natural Number Powers in a Field, (s+t)4=(s+t)3(s+t)(s+t)^{4}=(s+t)^{3}(s+t), so by Step 1 and distributivity

(s+t)4=(s3s+(3s2t)s+(3st2)s+t3s)+(s3t+(3s2t)t+(3st2)t+t3t).(s+t)^{4}=\bigl(s^{3}s+(3s^{2}t)s+(3st^{2})s+t^{3}s\bigr)+\bigl(s^{3}t+(3s^{2}t)t+(3st^{2})t+t^{3}t\bigr).

By associativity and commutativity of multiplication,

s3s=s4,(3s2t)s=3s3t,(3st2)s=3s2t2,t3s=st3,(3s2t)t=3s2t2,(3st2)t=3st3,t3t=t4.s^{3}s=s^{4},\quad (3s^{2}t)s=3s^{3}t,\quad (3st^{2})s=3s^{2}t^{2},\quad t^{3}s=st^{3},\quad (3s^{2}t)t=3s^{2}t^{2},\quad (3st^{2})t=3st^{3},\quad t^{3}t=t^{4}.

Collecting by associativity and commutativity of addition and using Step 0 in the forms s3t+3s3t=4s3ts^{3}t+3s^{3}t=4s^{3}t, 3s2t2+3s2t2=6s2t23s^{2}t^{2}+3s^{2}t^{2}=6s^{2}t^{2} and 3st3+st3=4st33st^{3}+st^{3}=4st^{3},

(s+t)4=s4+4s3t+6s2t2+4st3+t4,(s+t)^{4}=s^{4}+4s^{3}t+6s^{2}t^{2}+4st^{3}+t^{4},

which is claim 2.

Please log in to copy this version.

Citations

Loading…

Dependency Graph

0 prerequisites

Prerequisites

Loading...

proof421dca55...421dca55-8b91-485e-ba5e-f4762489f320

Target theorem

lemmaThe Expansion of the Square and the Fourth Power of a Sum of Real Numberslem:fourth-power-binomial-real-2026a

Comments

Loading…