### A Boundary Value Problem for a Second-Order Singular by Larin A. A.

• February 26, 2017
• Mathematics
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By Larin A. A.

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Additional resources for A Boundary Value Problem for a Second-Order Singular Elliptic Equation in a Sector on the Plane

Example text

1. There are an inﬁnite number of irrational numbers. Proof. 1 tells us that each √ these √ prime √ numbers has an √ of irrational square root. It follows that the list 2, 3, 5, 7, . . has no largest member and that’s what we set out to prove. We have shown that there are already inﬁnitely many irrational numbers that √ take the form p where p is a prime number. In fact there are many more irrational numbers than this and later on in, Chapter 7, we will look at proving the irrationality of some very famous numbers such as e – the base of natural logarithms and π – the ratio of the circumference of any circle to its diameter.

Now either b > |a| or b ≤ |a|. If b > |a|, |a + b| = b − |a| ≤ b + |a| = |a| + |b|. If b ≤ |a|, |a + b| = |a| − b ≤ |a| + |b|. (iv) a > 0, b ≤ 0. This is proved the same way as in (iii). 4) holds in all four cases. There are no other possibilities to consider and that completes the proof. 1. Now since the square of a number is always nonnegative it follows that |a + b|2 = (a + b)2 = a2 + 2ab + b2 = |a|2 + 2ab + |b|2 . g. 5) we get ab ≤ |ab| = |a||b|, and so we see that |a + b|2 ≤ |a|2 + 2|a||b| + |b|2 = (|a + b|)2 .

The essential difference between the discrete and the continuous is captured by the irrationals and this is one of the reasons why we ﬁnd them so strange. They are the ﬁrst type of number to take us away from our direct experience of the world around us. 4 A First Look at Infinity The real numbers are all ﬁnite numbers. No matter how difﬁcult it may be to pin down an irrational number through its decimal expansion, it still represents a ﬁxed point on the number line that measures a ﬁnite distance from the origin.