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Chapter 3: Problem 9

Is the following statement true or false? For all positive real numbers \(x\) and \(y, \sqrt{x+y} \leq \sqrt{x}+\sqrt{y}\).

Short Answer

Expert verified
The statement is true. For all positive real numbers \(x\) and \(y\), the inequality \(\sqrt{x+y} \leq \sqrt{x}+\sqrt{y}\) holds.

Step by step solution

01

Analyze the given inequality

We have the inequality \(\sqrt{x+y} \leq \sqrt{x}+\sqrt{y}\) to test for all positive real numbers x and y. Let's first note that both sides of the inequality are non-negative since x and y are positive real numbers, and square root functions always return non-negative results.
02

Square both sides of the inequality

Since both sides of the inequality are non-negative, we can square both sides of the inequality without flipping the direction of the inequality sign. This gives us: \((\sqrt{x+y})^2 \leq (\sqrt{x}+\sqrt{y})^2\). Simplifying both sides, we get: \(x+y \leq x+2\sqrt{x}\sqrt{y}+y\).
03

Isolate the term involving both x and y

Now we will isolate the term involving both x and y by subtracting x and y from both sides of the inequality: \[ 0 \leq 2\sqrt{x}\sqrt{y}. \]
04

Conclude the validity of the statement

Since the inequality \(0 \leq 2\sqrt{x}\sqrt{y}\) holds for all positive real numbers x and y (as both x and y are positive and thus their product is positive), we can conclude that the original statement, \(\sqrt{x+y} \leq \sqrt{x}+\sqrt{y}\), is indeed true for all positive real numbers x and y.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Square Roots
Square roots are mathematical operations that help us find a number which, when multiplied by itself, equals the original number we started with. For example, the square root of 9 is 3, because when 3 is multiplied by itself, we get 9. Square roots apply to all non-negative numbers, including zero.

A key property of square roots is that they always produce non-negative outputs. This is because they are defined in such a way that only the principal (non-negative) root is considered for real numbers.
  • The square root of zero is zero.
  • The square root of any positive number is also positive.
This characteristic is particularly useful when analyzing and comparing inequalities involving square roots, as seen in the context of our original exercise.
Inequality Proof
An inequality proof is a systematic way to demonstrate that an inequality holds true under certain conditions. In the context of the exercise, we aimed to prove whether the statement \( \sqrt{x+y} \leq \sqrt{x}+\sqrt{y} \) was valid for all positive real numbers.

To do this, we made use of some important strategies:
  • Squaring both sides: If both sides of an inequality are non-negative, we can square both sides without changing the inequality's direction. This step simplifies the process by removing the square roots, making it easier to compare the resulting expressions.
  • Isolating terms: By isolating specific terms, such as \(2\sqrt{x}\sqrt{y}\) in our inequality, we can evaluate whether the remaining statement is always true.
Ultimately, our steps confirmed the inequality, demonstrating it holds for all positive real numbers \(x\) and \(y\). Understanding this process is invaluable when working with similar mathematical expressions and proofs.
Positive Real Numbers
Positive real numbers are the set of all numbers greater than zero. They include numbers like 1, 2, 3.5, 0.001, and so forth.
  • They do not include zero, as zero is neither positive nor negative.
  • They also do not include negative numbers or imaginary numbers.
In mathematics, working with positive real numbers ensures that operations such as square roots do not encounter undefined or problematic results.

These numbers play a crucial role in many mathematical statements and proofs, including the one from our exercise. Since \(x\) and \(y\) are both positive real numbers in the exercise, we are guaranteed that their sums and products will also be positive, which simplifies the verification of inequalities, such as \(\sqrt{x+y}\) and \(\sqrt{x} + \sqrt{y}\), without fear of encountering negative or indeterminate forms.

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Most popular questions from this chapter

We have already seen examples of Pythagorean triples, which are natural numbers \(a, b,\) and \(c\) where \(a^{2}+\) \(b^{2}=c^{2}\). For example, \(3,4,\) and 5 form a Pythagorean triple as do \(5,\) \(12,\) and \(13 .\) One of the famous mathematicians of the 17 th century was Pierre de Fermat \((1601-1665)\). Fermat made an assertion that for each natural number \(n\) with \(n \geq 3,\) there are no positive integers \(a, b,\) and \(c\) for which \(a^{n}+b^{n}=c^{n}\). This assertion was discovered in a margin of one of Fermat's books after his death, but Fermat provided no proof. He did, however, state that he had discovered a truly remarkable proof but the margin did not contain enough room for the proof. This assertion became known as Fermat's Last Theorem but it more properly should have been called Fermat's Last Conjecture. Despite the efforts of mathematicians, this "theorem" remained unproved until Andrew Wiles, a British mathematician, first announced a proof in June of \(1993 .\) However, it was soon recognized that this proof had a serious gap, but a widely accepted version of the proof was published by Wiles in \(1995 .\) Wiles' proof uses many concepts and techniques that were unknown at the time of Fermat. We cannot discuss the proof here, but we will explore and prove the following proposition, which is a (very) special case of Fermat's Last Theorem. Proposition. There do not exist prime numbers \(a, b,\) and \(c\) such that \(a^{3}+b^{3}=c^{3}\) Although Fermat's Last Theorem implies this proposition is true, we will use a proof by contradiction to prove this proposition. For a proof by contradiction, we assume that there exist prime numbers \(a, b,\) and \(c\) such that \(a^{3}+b^{3}=c^{3}\). Since 2 is the only even prime number, we will use the following cases: (1) \(a=b=2 ;\) (2) \(a\) and \(b\) are both odd; and (3) one of \(a\) and \(b\) is odd and the other one is 2 . (a) Show that the case where \(a=b=2\) leads to a contradiction and hence, this case is not possible. (b) Show that the case where \(a\) and \(b\) are both odd leads to a contradiction and hence, this case is not possible. (c) We now know that one of \(a\) or \(b\) must be equal to 2 . So we assume that \(b=2\) and that \(a\) is an odd prime. Substitute \(b=2\) into the equation \(b^{3}=c^{3}-a^{3}\) and then factor the expression \(c^{3}-a^{3}\). Use this to obtain a contradiction. (d) Write a complete proof of the proposition.

(a) Prove that for each integer \(a\), if \(a \neq 0(\bmod 7),\) then \(a^{2} \neq 0(\bmod 7)\). (b) Prove that for each integer \(a\), if 7 divides \(a^{2}\), then 7 divides \(a\). (c) Prove that the real number \(\sqrt{7}\) is an irrational number.

Prove each of the following: (a) For each nonzero real number \(x,\left|x^{-1}\right|=\frac{1}{|x|}\). (b) For all real numbers \(x\) and \(y,|x-y| \geq|x|-|y|\) Hint: An idea that is often used by mathematicians is to add 0 to an expression "intelligently". In this case, we know that \((-y)+y=0\). Start by adding this "version" of 0 inside the absolute value sign of \(|x|\). (c) For all real numbers \(x\) and \(y, \| x|-| y|| \leq|x-y|\).

See the instructions for Exercise (19) on page 100 from Section 3.1 . (a) For all nonzero integers \(a\) and \(b,\) if \(a+2 b \neq 3\) and \(9 a+2 b \neq 1,\) then the equation \(a x^{3}+2 b x=3\) does not have a solution that is a natural number. \mathrm{P} \text { Proof } . We will prove the contrapositive, which is For all nonzero integers \(a\) and \(b\), if the equation \(a x^{3}+2 b x=3\) has a solution that is a natural number, then \(a+2 b=3\) or \(9 a+2 b=1\). So we let \(a\) and \(b\) be nonzero integers and assume that the natural number \(n\) is a solution of the equation \(a x^{3}+2 b x=3\). So we have $$ \begin{aligned} a n^{3}+2 b n &=3 \quad \text { or } \\ n\left(a n^{2}+2 b\right) &=3 \end{aligned} $$ So we can conclude that \(n=3\) and \(a n^{2}+2 b=1\). Since we now have the value of \(n\), we can substitute it in the equation \(a n^{3}+2 b n=3\) and obtain \(27 a+6 b=3\). Dividing both sides of this equation by 3 shows that \(9 a+2 b=1\). So there is no need for us to go any further, and this concludes the proof of the contrapositive of the proposition. (b) For all nonzero integers \(a\) and \(b,\) if \(a+2 b \neq 3\) and \(9 a+2 b \neq 1,\) then the equation \(a x^{3}+2 b x=3\) does not have a solution that is a natural number. We will use a proof by contradiction. Let us assume that there exist nonzero integers \(a\) and \(b\) such that \(a+2 b=3\) and \(9 a+2 b=1\) and \(a n^{3}+2 b n=3,\) where \(n\) is a natural number. First, we will solve one equation for \(2 b\); doing this, we obtain $$ \begin{aligned} a+2 b &=3 \\ 2 b &=3-a . \end{aligned} $$ We can now substitute for \(2 b\) in \(a n^{3}+2 b n=3\). This gives $$ \begin{array}{l} a n^{3}+(3-a) n=3 \\ a n^{3}+3 n-a n=3 \\ n\left(a n^{2}+3-a\right)=3 . \end{array} $$ By the closure properties of the integers, \(\left(a n^{2}+3-a\right)\) is an integer and, hence, equation (2) implies that \(n\) divides \(3 .\) So \(n=1\) or \(n=3\). When we substitute \(n=1\) into the equation \(a n^{3}+2 b n=3,\) we obtain \(a+2 b=3\). This is a contradiction since we are told in the proposition that \(a+2 b \neq 3\). This proves that the negation of the proposition is false and, hence, the proposition is true.

Recall that a Pythagorean triple consists of three natural numbers \(a, b,\) and \(c\) such that \(a
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