G.371 unequal magnitudes set

X.1

Two unequal magnitudes being set out, if from the greater there be subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process be repeated continually, there is left some magnitude which is less than the lesser magnitude set out.
unequal_magnitudes_set(/ A B /, C)
given / A B /, C be two unequal magnitudes of which / A B / is the greater : I say that, if from / A B / there be subtracted a magnitude greater than its half, from that which is left a magnitude greater than its half, if this process be repeated continually, there will be left some magnitude which will be less than the magnitude C
prove The theorem can be similarly proved even if the parts subtracted be halves
A K H B C D F G E
given / A B /, C be two unequal magnitudes of which / A B / is the greater : I say that, if from / A B / there be subtracted a magnitude greater than its half, from that which is left a magnitude greater than its half, if this process be repeated continually, there will be left some magnitude which will be less than the magnitude C
∵ magnitudes_said_ratio_def_v_def_4assert C if multiplied will sometime be greater than / A B /
assert Let it be multiplied, and let / D E / be a multiple of C, and greater than / A B /
construct Let / D E / be divided into the parts / D F /, / F G /, / G E / equal to C, from / A B / let there be subtracted / B H / greater than its half, and, from / A H /, / H K / greater than its half, and let this process be repeated continually until the divisions in / A B / are equal in multitude with the divisions in / D E /.
assert Let, then, / A K /, / K H /, / H B / be divisions which are equal in multitude with / D F /, / F G /, / G E /
assert / D E / > / A B /, and from / D E / there has been subtracted / E G / less than its half, and, from / A B /, / B H / greater than its half,
assert / G D / > / H A /
assert since / G D / > / H A /, and there has been subtracted, from / G D /, the half / G F /, and, from / H A /, / H K / greater than its half,
assert / D F / > / A K /
assert / D F / = C
assert C is also greater than / A K /
assert / A K / < C
∴ The theorem can be similarly proved even if the parts subtracted be halves

Dependency Graph

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Required for

G.556, G.557, G.558, G.559, G.563, G.565, G.570, G.571, G.572, G.573, G.574, G.577, G.578