DG5, DG6 - Identity and inverses

DG5, DG6 - Identity and inverses

An identity leaves every element unchanged; an inverse combines with an element to return that identity. You will find and prove identities, check whether inverses belong to the given set, and keep left and right calculations separate when order matters.

Two-sided identity

For the operation a∗b=a+b+2a*b=a+b+2, what plays the role that zero plays in ordinary addition? Trying zero gives a∗0=a+2a*0=a+2, so the familiar answer fails. We need an element that cancels the extra 22.

Identity element

An identity for an operation ∗* on SS is an element e∈Se\in S such that

e∗a=aanda∗e=afor every a∈S.e*a=a\quad\text{and}\quad a*e=a\qquad\text{for every }a\in S.

The identity must work on both sides and for every element of the stated set.

Ordinary addition has identity 00; ordinary multiplication has identity 11. For matrix multiplication on all real 2×22\times2 matrices, the identity is I=(1001)\mathbf I=\begin{pmatrix}1&0\\0&1\end{pmatrix}, since IA=AI=A\mathbf I\mathbf A=\mathbf A\mathbf I=\mathbf A. The word “identity” describes its job, not its numerical value.

Worked example 1: finding and proving an identity

Question. For a∗b=a+b+2a*b=a+b+2 on Z\mathbb Z, find the identity and prove it works.

Plan. Solve the unchanged-output condition, then check the candidate on both sides for arbitrary aa.

a∗e=a⟹a+e+2=a⟹e=−2.a*e=a\quad\Longrightarrow\quad a+e+2=a\quad\Longrightarrow\quad e=-2.

The candidate belongs to Z\mathbb Z. For every integer aa,

a∗(−2)=a−2+2=a,(−2)∗a=−2+a+2=a.a*(-2)=a-2+2=a,\qquad(-2)*a=-2+a+2=a.

Answer. The identity is −2-2, with both required equations proved.

Check. For the boundary input a=−2a=-2, (−2)∗(−2)=−2(-2)*(-2)=-2: even the identity itself is left unchanged.

Keep the rule fixed, but change only its set to the nonnegative integers. The calculations still make sense and are closed; the candidate's membership changes.

Worked example 2: a candidate outside the set

Question. Does a∗b=a+b+2a*b=a+b+2 have an identity on S={0,1,2,…}S=\{0,1,2,\ldots\}?

Plan. Derive the necessary candidate and test whether it is allowed.

Every sum a+b+2a+b+2 is a nonnegative integer, so the operation is closed. But any identity would have to satisfy

0∗e=0⟹e+2=0⟹e=−2.0*e=0\quad\Longrightarrow\quad e+2=0\quad\Longrightarrow\quad e=-2.

Answer. There is no identity in SS, because the only possible candidate −2-2 is excluded.

Check. For every allowed e≥0e\ge0, 0∗e=e+2≥20*e=e+2\ge2, so no allowed element even leaves 00 unchanged.

In a finite table, an identity's row must reproduce the column headings in their existing order, and its column must reproduce the row headings. Merely containing every element once is weaker: the positions matter.

Watch out

Solving a∗e=aa*e=a produces a candidate. Finish by checking membership and both identities for every aa; one numerical trial is insufficient.

Use a new constant to find the corresponding identity.

Why the identity is unique

Finding one identity settles the question completely: an operation cannot have two different two-sided identities. The proof is short because each proposed identity must leave the other unchanged.

Suppose ee and ff are both identities in the same set under the same operation. Since ee is an identity, it leaves ff unchanged when placed on the left. Since ff is an identity, it leaves ee unchanged when placed on the right. Thus

e∗f=fande∗f=e⟹e=f.e*f=f\quad\text{and}\quad e*f=e\quad\Longrightarrow\quad e=f.

No associativity or commutativity was assumed. The same expression e∗fe*f has been evaluated in two legitimate ways.

Uniqueness

An object is unique when any two objects satisfying its defining conditions must be equal. Proving uniqueness does not itself prove existence: an operation may have no identity.

For example, multiplication on {2,3,4,…}\{2,3,4,\ldots\} is closed but has no identity. Any identity would need 2e=22e=2, forcing e=1e=1, outside the set. The uniqueness theorem remains true: it says there cannot be two identities, not that one must exist.

The proof also works if ee is only a left identity and ff only a right identity: e∗f=fe*f=f by the left property and e∗f=ee*f=e by the right property, so they are equal and that common element is two-sided. In contrast, two elements that are both merely left identities need not be equal; their defining property gives no reason for e∗f=ee*f=e.

Key idea

Existence requires an allowed element that works. Uniqueness requires proving that any two candidates that work must coincide. These are different claims.

Use the argument itself, without assuming any extra operation law.

Inverses within the given set

An inverse undoes an element using the chosen operation. Its target is the operation's identity, which need not be zero.

Inverse element

For an operation on SS with two-sided identity ee, a two-sided inverse of a∈Sa\in S is an element b∈Sb\in S satisfying

a∗b=eandb∗a=e.a*b=e\quad\text{and}\quad b*a=e.

The notation a−1a^{-1} means inverse under the specified operation. It means 1/a1/a only when that operation is ordinary multiplication.

Under ordinary addition on Z\mathbb Z, the identity is 00 and the inverse of aa is −a-a. Under +n+_n on the full residue set, the inverse of a≠0a\ne0 is n−an-a; the inverse of 00 is 00. Under multiplication, a zero element can never reach the identity 11, since 0b=00b=0 for every bb.

Worked example 3: an inverse missing from the set

Question. Does 33 have an inverse under ordinary multiplication on Z\mathbb Z?

Plan. Solve for the element that would give the identity 11, then check membership.

3b=1⟹b=13.3b=1\quad\Longrightarrow\quad b=\frac13.

Answer. No inverse exists in Z\mathbb Z, because 13\frac13 is not an integer. Multiplication on Z\mathbb Z is closed and has identity 11, but that does not give every element an inverse.

Check. Every integer multiple of 33 is divisible by 33; 11 is not, so an integer solution is impossible.

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