Exercise 1: Why a Word in Every Document Gets IDF Exactly Zero — Possible Solution
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THE IDF FORMULA, PER THIS CHAPTER
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Per this chapter:
  IDF(word) = log( total documents / documents containing word )

WORKING THROUGH A WORD IN EVERY DOCUMENT
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If a word appears in every single document in the corpus, then "documents
containing word" equals "total documents" — the two numbers in the
formula are identical. Substituting this into the formula:

  IDF(word) = log( total documents / total documents ) = log(1)

WHY log(1) EQUALS EXACTLY ZERO
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The logarithm of 1 is 0 for any valid logarithmic base — this is a
basic property of logarithms (any base raised to the power of 0 equals
1, so the logarithm of 1 is always 0, regardless of which base is
used). There's nothing approximate or asymptotic about this — dividing
identical numbers always produces exactly 1, and the logarithm of
exactly 1 is always exactly 0.

WHAT THIS MEANS FOR THE OVERALL TF-IDF SCORE
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Per this chapter, "TF-IDF = TF × IDF." If IDF is exactly 0, then TF-IDF
= TF × 0 = 0, regardless of what TF's own value happens to be. Even a
word that appears extremely frequently within one specific document —
a high TF value — still produces a final TF-IDF score of exactly zero
once multiplied by an IDF of zero.

WHY THIS MAKES SENSE GIVEN WHAT TF-IDF IS TRYING TO MEASURE
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Per this chapter, TF-IDF scores high specifically for words that are
"frequent in this document and rare across the corpus overall —
genuinely distinctive for this specific document." A word appearing in
literally every document, by definition, cannot distinguish any one
document from any other — there's no document it's absent from to
contrast against. Assigning it a TF-IDF score of exactly zero
correctly reflects that this word carries zero distinguishing power
for identifying what makes any particular document unique, no matter
how often it's repeated within that document.

WHY THIS WORKS AS AN ANSWER
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It substitutes the specific case (a word in every document) directly
into the chapter's own IDF formula, shows algebraically why this
produces log(1) = 0, and explains why a zero IDF forces the entire
TF-IDF score to zero regardless of TF, in a way that correctly reflects
the word's own genuine lack of distinguishing power.