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Tasaio
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Spivak's Calculus, 5(x) -- "Use (ix) backwards..."
Prove the following:
(x) If a,b\geq0 and a^{2}<b^{2}, then a<b. (Use (ix), backwards.)
(ix) If 0 \leq a<b, then a^{2}<b^{2}.
Suppose a,b\geq0 and a^{2}<b^{2}.
Here's my problem. What does "Use (ix) backwards" mean? I'll assume he means to use the converse of (ix). In that case...
The converse of (ix):
\neg(a^{2}<b^{2})\rightarrow\neg(0\leq a<b)
Hence (a^{2}\geq b^{2})\rightarrow\neg(0\leq a\&\&a<b);
hence (a^{2}\geq b^{2})\rightarrow(0>a)\Vert(a\geq b). (\star)
Since a^{2}<b^{2}, then a^{2} \leq b^{2}. So b^{2} \geq a^{2}.
Then by (\star), (0>b)\Vert(b \geq a).
Since b \geq 0, then we know 0>b cannot be true.
This means that b\geq a must be true.
But if b=a, then b^{2}=a^{2}; this is a contradiction since we are given that a^{2}<b^{2}.
Hence b>a must be true.
Hence a<b.
Homework Statement
Prove the following:
(x) If a,b\geq0 and a^{2}<b^{2}, then a<b. (Use (ix), backwards.)
Homework Equations
(ix) If 0 \leq a<b, then a^{2}<b^{2}.
The Attempt at a Solution
Suppose a,b\geq0 and a^{2}<b^{2}.
Here's my problem. What does "Use (ix) backwards" mean? I'll assume he means to use the converse of (ix). In that case...
The converse of (ix):
\neg(a^{2}<b^{2})\rightarrow\neg(0\leq a<b)
Hence (a^{2}\geq b^{2})\rightarrow\neg(0\leq a\&\&a<b);
hence (a^{2}\geq b^{2})\rightarrow(0>a)\Vert(a\geq b). (\star)
Since a^{2}<b^{2}, then a^{2} \leq b^{2}. So b^{2} \geq a^{2}.
Then by (\star), (0>b)\Vert(b \geq a).
Since b \geq 0, then we know 0>b cannot be true.
This means that b\geq a must be true.
But if b=a, then b^{2}=a^{2}; this is a contradiction since we are given that a^{2}<b^{2}.
Hence b>a must be true.
Hence a<b.
