Proof
Construct with commensurable in length with the
assigned rational and the square-discriminant commensurable with the
greater (X.29).
Knowledge graph · drag to pan, scroll to zoom, click a node to navigate
Full neighborhood
Depends on (3)
- X.29Proposition X.29To find two rational straight lines commensurable in square only such that the square on the greater is greater than…
- X.36Proposition X.36If two rational straight lines commensurable in square only be added together, the whole is irrational; and let it be…
- X.II.1Definition X.II.1Given a rational straight line and a binomial, divided into its terms, let the square of the greater term be greater…
Required by (dependents) (7)
- X.49Proposition X.49To find the second binomial straight line.
- X.50Proposition X.50To find the third binomial straight line.
- X.51Proposition X.51To find the fourth binomial straight line.
- X.54Proposition X.54If an area be contained by a rational straight line and the first binomial, the side of the area is the irrational…
- X.60Proposition X.60The square on the binomial straight line applied to a rational straight line produces as breadth the first binomial.
- X.66Proposition X.66A straight line commensurable in length with a binomial straight line is itself also binomial and the same in order.
- X.85Proposition X.85To find the first apotome.
Discussion
No replications, contradictions, or comments registered yet for this claim.
Replicate or annotate this claim
Replicate to register a fresh attempt; contradict, extend, or comment otherwise. Authors can post a claim-retraction with the reason taxonomy from RRP-0020.
Sign in with ORCID to annotate this claim.