- #1
Saladsamurai
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Hello all!
I am (painfully) going through the first chapter of Spivak's Calculus. At one point he introduces the property: if a, b, and c are any numbers, then [itex]a\cdot(b+c) = a\cdot b +a\cdot c[/itex]. He then uses this property in an example in which he shows that the only time that [itex] a - b = b- a[/itex] is when a = b.[itex]a - b = b - a [/itex]
[itex] \Rightarrow (a - b) + b = (b - a) + b = b + (b - a) [/itex]
[itex]\Rightarrow a = b + b - a [/itex]
[itex]\Rightarrow a + a = (b + b - a) + a = b + b [/itex]
[itex]\mathbf{ \Rightarrow a \cdot (1 + 1) = b \cdot (1 + 1) } [/itex]
[itex]\Rightarrow a = b [/itex]I am unsure of why the step in bold was necessary? Can someone elaborate as to why he included this step? In fact, he said it was necessary to include this step.
I am (painfully) going through the first chapter of Spivak's Calculus. At one point he introduces the property: if a, b, and c are any numbers, then [itex]a\cdot(b+c) = a\cdot b +a\cdot c[/itex]. He then uses this property in an example in which he shows that the only time that [itex] a - b = b- a[/itex] is when a = b.[itex]a - b = b - a [/itex]
[itex] \Rightarrow (a - b) + b = (b - a) + b = b + (b - a) [/itex]
[itex]\Rightarrow a = b + b - a [/itex]
[itex]\Rightarrow a + a = (b + b - a) + a = b + b [/itex]
[itex]\mathbf{ \Rightarrow a \cdot (1 + 1) = b \cdot (1 + 1) } [/itex]
[itex]\Rightarrow a = b [/itex]I am unsure of why the step in bold was necessary? Can someone elaborate as to why he included this step? In fact, he said it was necessary to include this step.
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