- #1
Aziza
- 190
- 1
"Prove that there exists an odd integer M such that for all real numbers r larger than
M, [itex]\frac{1}{2r} < \frac{1}{100}[/itex]"
How to express this using logical symbols?
Literally I understand this statement as:
[itex](\exists M\in [/itex][PLAIN]http://upload.wikimedia.org/wikipedia/en/math/3/f/3/3f3c78f02a9c53f5460f4bcc2e7dd3cb.png[itex])[([/itex]M [Broken] is odd[itex])\wedge(\forall r\inℝ)(r>M\wedge\frac{1}{2r} < \frac{1}{100})][/itex]
But all r in ℝ cannot be greater than some M. There will always be r<=M if the only restriction on r is that it is in ℝ. So this statement can't be proven true. But would it be wrong to interpret the statement as follows:
[itex](\exists M\in [/itex][PLAIN]http://upload.wikimedia.org/wikipedia/en/math/3/f/3/3f3c78f02a9c53f5460f4bcc2e7dd3cb.png[itex])[([/itex]M [Broken] is odd[itex])\wedge(\forall r\inℝ)(r>M\Rightarrow\frac{1}{2r} < \frac{1}{100})][/itex]
(the last "and" changed to "imples")...then M=51 would prove the statement true...
M, [itex]\frac{1}{2r} < \frac{1}{100}[/itex]"
How to express this using logical symbols?
Literally I understand this statement as:
[itex](\exists M\in [/itex][PLAIN]http://upload.wikimedia.org/wikipedia/en/math/3/f/3/3f3c78f02a9c53f5460f4bcc2e7dd3cb.png[itex])[([/itex]M [Broken] is odd[itex])\wedge(\forall r\inℝ)(r>M\wedge\frac{1}{2r} < \frac{1}{100})][/itex]
But all r in ℝ cannot be greater than some M. There will always be r<=M if the only restriction on r is that it is in ℝ. So this statement can't be proven true. But would it be wrong to interpret the statement as follows:
[itex](\exists M\in [/itex][PLAIN]http://upload.wikimedia.org/wikipedia/en/math/3/f/3/3f3c78f02a9c53f5460f4bcc2e7dd3cb.png[itex])[([/itex]M [Broken] is odd[itex])\wedge(\forall r\inℝ)(r>M\Rightarrow\frac{1}{2r} < \frac{1}{100})][/itex]
(the last "and" changed to "imples")...then M=51 would prove the statement true...
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