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Consider the following statement. For all positive real numbers r and s, Vr+s # Vr + Vs. Construct a proof by contradiction for the statement by selecting sentences from the following scrambled list and putting them in the correct order. By the zero product property, at least one of vr or Vs equals 0, which implies that r or s equals 0. But this is a contradiction because r and s are positive. Simplifying the equation gives that 0 2rs. Simplifying the equation gives that 0 = 2yrys. Squaring both sides of the equation gives that r +s = r + 2yrys + s. By the zero product property, at least one of r or s equals 0. Squaring both sides of the equation gives that r + s =r + 2rs + s. Proof by contradiction: 1. Suppose not. That is, suppose there exists positive real numbers r and s such that Vr + s = Vr + vs. 2. By the zero product property, at least one of v(r) or V(s) equals 0, which implies that r or s equals 0. v Select-- 3. 4. Select--- 5. Select- 6. Thus, we have reached a contradiction and have proved the statement.

Consider the following statement. For all positive real numbers r and s, Vr+s # Vr + Vs. Construct a proof by contradiction for the statement by selecting sentences from the following scrambled list and putting them in the correct order. By the zero product property, at least one of vr or Vs equals 0, which implies that r or s equals 0. But this is a contradiction because r and s are positive. Simplifying the equation gives that 0 2rs. Simplifying the equation gives that 0 = 2yrys. Squaring both sides of the equation gives that r +s = r + 2yrys + s. By the zero product property, at least one of r or s equals 0. Squaring both sides of the equation gives that r + s =r + 2rs + s. Proof by contradiction: 1. Suppose not. That is, suppose there exists positive real numbers r and s such that Vr + s = Vr + vs. 2. By the zero product property, at least one of v(r) or V(s) equals 0, which implies that r or s equals 0. v Select-- 3. 4. Select--- 5. Select- 6. Thus, we have reached a contradiction and have proved the statement.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Consider the following statement. For all positive real numbers r and s, Vr + s # Vr + Vs. Construct a proof by contradiction for the statement by selecting sentences from the following scrambled list and putting them in the correct order. By the zero product property, at least one of vr or Vs equals 0, which implies that r or s equals 0. But this is a contradiction because r and s are positive. Simplifying the equation gives that 0 2rs. Simplifying the equation gives that 0 = 2yrys. Squaring both sides of the equation gives that r + s = r + 2VTVS+ s. By the zero product property, at least one of r or s equals 0. Squaring both sides of the equation gives that r + s =r + 2rs + s. Proof by contradiction: 1. Suppose not. That is, suppose there exists positive real numbers r and s such that Vr + s = Vr + vs. 2. By the zero product property, at least one of v(r) or V(s) equals 0, which implies that r or s equals 0. v Select-- 3. 4. Select--- 5. Select- 6. Thus, we have reached a contradiction and have proved the statement.
Transcribed Image Text:Consider the following statement. For all positive real numbers r and s, Vr + s # Vr + Vs. Construct a proof by contradiction for the statement by selecting sentences from the following scrambled list and putting them in the correct order. By the zero product property, at least one of vr or Vs equals 0, which implies that r or s equals 0. But this is a contradiction because r and s are positive. Simplifying the equation gives that 0 2rs. Simplifying the equation gives that 0 = 2yrys. Squaring both sides of the equation gives that r + s = r + 2VTVS+ s. By the zero product property, at least one of r or s equals 0. Squaring both sides of the equation gives that r + s =r + 2rs + s. Proof by contradiction: 1. Suppose not. That is, suppose there exists positive real numbers r and s such that Vr + s = Vr + vs. 2. By the zero product property, at least one of v(r) or V(s) equals 0, which implies that r or s equals 0. v Select-- 3. 4. Select--- 5. Select- 6. Thus, we have reached a contradiction and have proved the statement.
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