Fill in the blanks in the following proof by contradiction that there is no least positive real number. Proof: Suppose npt. That is, suppose that there is a least positive real number x . [We must deduce ___(a)___] Consider the number x / 2. Since x is a positive real numbers x / 2 is also __(b)__ In addition, we can deduce that x / 2 < x by __(c)___ and dividing __(d)__. Hence x / 2 is a positive real number that is less than the least positive real number. This is a _(e)__. [Thus the supposition id false, and so there is no least positive real number.]
Fill in the blanks in the following proof by contradiction that there is no least positive real number. Proof: Suppose npt. That is, suppose that there is a least positive real number x . [We must deduce ___(a)___] Consider the number x / 2. Since x is a positive real numbers x / 2 is also __(b)__ In addition, we can deduce that x / 2 < x by __(c)___ and dividing __(d)__. Hence x / 2 is a positive real number that is less than the least positive real number. This is a _(e)__. [Thus the supposition id false, and so there is no least positive real number.]
Solution Summary: The author explains the supposition that there is no least positive real number by contradiction.
Fill in the blanks in the following proof by contradiction that there is no least positive real number. Proof: Suppose npt. That is, suppose that there is a least positive real number x. [We must deduce ___(a)___] Consider the number
x
/
2.
Since x is a positive real numbers
x
/
2
is also __(b)__ In addition, we can deduce that
x
/
2
<
x
by __(c)___ and dividing __(d)__. Hence
x
/
2
is a positive real number that is less than the least positive real number. This is a _(e)__. [Thus the supposition id false, and so there is no least positive real number.]
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MFCS unit-1 || Part:1 || JNTU || Well formed formula || propositional calculus || truth tables; Author: Learn with Smily;https://www.youtube.com/watch?v=XV15Q4mCcHc;License: Standard YouTube License, CC-BY