Question

Find the flaw in the following "proof": Let $a$ and $b$ be real numbers such that $a=b .$ Then $a b=b^{2}$ Therefore, $a^{2}-a b=a^{2}-b^{2}$ Factoring, $a(a-b)=(a+b)(a-b)$ Cancel $a-b$ from both sides: $$a=a+b$$ since $a=b,$ this yields $a=2 a$ Cancel a from both sides. Then we get $1=2$ . Let $a, b,$ and $c$ be any real numbers. Then $a<b$ if and only if there is a positive real number $x$ such that $a+x=b$ . Use this fact to prove each.

Name: Problem 39, Chapter 1: Discrete Mathematics with Applications
Uploaded: 2021-08-26T10:46:36-08:00
Duration: 1 min 51 s
Channel: Mohamed Mohamed

   Find the flaw in the following "proof":
Let $a$ and $b$ be real numbers such that $a=b .$ Then $a b=b^{2}$
Therefore, $a^{2}-a b=a^{2}-b^{2}$
Factoring, $a(a-b)=(a+b)(a-b)$
Cancel $a-b$ from both sides:
$$a=a+b$$
since $a=b,$ this yields $a=2 a$
Cancel a from both sides.
Then we get $1=2$ .
Let $a, b,$ and $c$ be any real numbers. Then $a<b$ if and only if there is a positive real number $x$ such that $a+x=b$ . Use this fact to prove each.

Discrete Mathematics with Applications

Thomas Koshy 1st Edition

Chapter 1, Problem 39

Instant Answer

Solved by Expert Mohamed Mohamed

08/26/2021

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This implies that $a-b=0$. Show more…

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Find the flaw in the following "proof": Let $a$ and $b$ be real numbers such that $a=b .$ Then $a b=b^{2}$ Therefore, $a^{2}-a b=a^{2}-b^{2}$ Factoring, $a(a-b)=(a+b)(a-b)$ Cancel $a-b$ from both sides: $$a=a+b$$ since $a=b,$ this yields $a=2 a$ Cancel a from both sides. Then we get $1=2$ . Let $a, b,$ and $c$ be any real numbers. Then $a<b$ if and only if there is a positive real number $x$ such that $a+x=b$ . Use this fact to prove each.

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Key Concepts

Division by Zero

Division by zero is undefined in mathematics and is a common pitfall in algebraic manipulations. In the proof, the step where the factor (a - b) is canceled is flawed because, when a equals b, the expression (a - b) equals zero. Canceling a zero factor invalidates the operation, leading to a logically incorrect conclusion.

Cancellation of Common Factors

When canceling a common factor from both sides of an equation, it is essential to ensure that the factor is not zero. In this proof, the cancellation step mistakenly assumes that (a - b) is a nonzero term, whereas it is actually equal to zero due to the initial condition a = b. This misuse of the cancellation property is the core mistake in the argument.

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