Indirect Argument: Contradiction and Contraposition
... Proposition 1: For any integer n, if n2 is even then n is also even. Proof: The contrapositive is: For any integer n, if n is not even then n2 is not even. Let’s show (1) by direct proof. Suppose n is not even. Then n is odd. So n=2k+1 for some kZ. Hence n2 =(2k+1)2=4k2+4k+1 Thus, n2 is not even. ...
... Proposition 1: For any integer n, if n2 is even then n is also even. Proof: The contrapositive is: For any integer n, if n is not even then n2 is not even. Let’s show (1) by direct proof. Suppose n is not even. Then n is odd. So n=2k+1 for some kZ. Hence n2 =(2k+1)2=4k2+4k+1 Thus, n2 is not even. ...
UProperty 1
... Without loss of generality, suppose the sequence is monotonically increasing. Since the sequence is a non-empty bounded set of real numbers, by the completeness property, it has a LUB, say, l. We claim that l is the limit of the sequence. For given any 0 , then l l , hence l cannot be ...
... Without loss of generality, suppose the sequence is monotonically increasing. Since the sequence is a non-empty bounded set of real numbers, by the completeness property, it has a LUB, say, l. We claim that l is the limit of the sequence. For given any 0 , then l l , hence l cannot be ...
1-4 Properties of Real Numbers
... – To add two numbers with different signs, subtract their absolute values. – The sum has the same sign as the addend with the greater absolute value. – Ex. -3 + 4 = 1 ...
... – To add two numbers with different signs, subtract their absolute values. – The sum has the same sign as the addend with the greater absolute value. – Ex. -3 + 4 = 1 ...
Unique representations of real numbers in non
... One of the consequences of this formula is that the function q 7→ dimH (Aq ) is continuous (this is because the function q 7→ {un (q)}∞ n=1 is continuous). However, it is not Hölder continuous, as dimH (Aq ) ≥ C/| log(q − qc )| when q > qc for some C > 0, n because, as is easy to see, qn − qc = O(q ...
... One of the consequences of this formula is that the function q 7→ dimH (Aq ) is continuous (this is because the function q 7→ {un (q)}∞ n=1 is continuous). However, it is not Hölder continuous, as dimH (Aq ) ≥ C/| log(q − qc )| when q > qc for some C > 0, n because, as is easy to see, qn − qc = O(q ...
Math 319/320 Homework 1
... (ii) All roses are red. (iii) Some real numbers do not have a square root. (iv) If you are rich and famous, you are happy. Problem 2. Provide a counterexample for each of the following statements: (i) For every real number x, if x2 > 4, then x > 2. (ii) For every positive integer n, n2 + n + 41 is a ...
... (ii) All roses are red. (iii) Some real numbers do not have a square root. (iv) If you are rich and famous, you are happy. Problem 2. Provide a counterexample for each of the following statements: (i) For every real number x, if x2 > 4, then x > 2. (ii) For every positive integer n, n2 + n + 41 is a ...
The Fundamental Theorem of Algebra
... The multiplicative property of radicals only works for positive values under the radical sign Instead use imaginary numbers ...
... The multiplicative property of radicals only works for positive values under the radical sign Instead use imaginary numbers ...
Full text
... for publication in the Quarterly should be sent to Verner E. Hoggatt, J r . , Mathematics Department, San Jose State College, San Jose, Calif. AH manuscripts should be typed, double-spaced. Drawings should be made the same size as they will appear in the Quarterly, and should be done in India ink on ...
... for publication in the Quarterly should be sent to Verner E. Hoggatt, J r . , Mathematics Department, San Jose State College, San Jose, Calif. AH manuscripts should be typed, double-spaced. Drawings should be made the same size as they will appear in the Quarterly, and should be done in India ink on ...
