Is the difference of two positive rational numbers always positive?
The word "difference" is not so well defined that an indisputable answer is possible. I would say that the difference of 4 and 7 is 3, the same as the difference of 7 and 4. Observing my usage, the difference of any two distinct real numbers is positive. So what about the expression 4 - 7? I would never call that the difference of 4 and 7. That is 4 minus 7.
Is the sum of two integers always greater than the difference between them? Why or why not? Give examples to support your answer.?
Sum{m and n} > Difference{m and n} I'll assume you mean the magnitude of the difference because it could be + or – depending on which is greater. For integers "m" and "n", it is TRUE ONLY if: m ≥ 1 and n ≥ 1 Examples for "m" and "n": 7 + (-7) = 0 and 7 – (-7) = 12 ... FALSE because: n ≤ 1 3 + 0 = 3 and 3 – 0 = 3 ... FALSE because: n ≤ 1 3 + 1 = 4 and 3 – 1 = 2 ... TRUE
Prove that among any given n + 1 positive integers, there are always two whose difference is divisible by n?
Let a1, a2, ..., an, an+1 be the numbers. For each number, calculate the remainder that results from dividing the number by n. So for example, a1 mod n gives a remainder of r1. a2 mod n gives a remainder of r2, etc. A remainder can only take the values of 0 through n-1 or a total of n values. But we are calculating a total of n+1 remainders. By the Pigeon Hole principle, there must be (at least) two of the remainders that are the same. So that means we will have two numbers such that: - The first is equal to an integer multiple of n plus the remainder --> jn + r - The second is equal to an integer multiple of n plus the remainder --> kn + r Subtracting these numbers, we have: jn + r - (kn + r) The remainders will cancel out leaving: jn - kn = (j - k)n This difference is therefore evenly divisible by n.
Prove that amongst any n + 1 arbitrary integers there are always two whose difference is divisible by n?
Use the pigeonhole principle to show that at least two of the arbitrary integers have the same remainder when divided by n. (show that at least two of them are congruent mod n). Then show that the difference between these two numbers is divisible by n.
Let take two natural integers, [math]x=\sum_{i\geq 0}x_i10^i[/math] and [math]y=\sum_{i\geq 0}y_i10^i[/math], written according their respective base-ten decomposition ([math]\forall i, x_i,y_i\in\{0,1,2,3,4,5,6,7,8,9\}[/math]).Then, [math]x+y = 10\sum_{i\geq 1}(x_i+y_i)10^{i-1}+(x_0+y_0)=x_0+y_0[10][/math]. (NB: [math]a=b[m][/math] stands for a is equal to b modulo m) And, similarly, [math]x-y=x_0-y_0[10][/math].Thus, it leads us to solve :“Given a set of seven numbers in S={0,1,2,3,4,5,6,7,8,9}, there is a pair whose sum or whose difference is a multiple of 10.”By the way, we lost the “distinct” aspect, for example, if we pick 42 and 72, which are distinct, but whose the last digit is the same.Let pick seven numbers among the set [math]S=\{0,1,2,3,4,5,6,7,8,9\}[/math].First case :we picked at least two non distinct (= equal) elements.The difference is 0 which is a multiple of 10.Second case : all our picked elements are distinct.Since we are only proving existence, we can forget a little the operation “x-y”.Here are all the pairs of distinct elements of S whose the sum is a multiple of 10 :P={a=(1,9) ; b=(2,8) ; c=(3,7) ; d=(4,6)} (the cardinal of P is 4)We actually picked 7, 6, or 5 elements from [math]S'=\{1,2,3,4,6,7,8,9\}[/math] , according if we picked {0}, {5}, both, or none.For each picked number, we can map it to its corresponding pair (1 is mapped to a, 2 to b, 3 to c, …, 9 to a). In other words, we built an application from S’ to P (finite sets), where the cardinal of S’ is strictly greater than P !We can now apply the pigeon holes result : it exists two distinct numbers in S, such that the pair belongs to P, which means the sum of these two elements is a multiple of 10.CONCLUSION : In a set of seven distinct positive integers, there is a pair whose sum or whose difference is a multiple of 10.
Prove that between any two different real numbers there is a rational number and an irrational number.?
L. E. Gant mentioned "and the limit is zero" When you take caculus, you will learn how the limit operator works, and later when you learn about computing sums of infinite series you will use the limit operator to find if a particular sum converges and what it converges to. The limit operator allows us to zing past the quandries you talk about. Infinity, and infinitesimals, are constructs of the human mind. The real world is not infinite, it has constraints on its size. Since the real world works, mathematics had to learn from the real world and create tools to explain why physics happens the way it does. Think of math as a bookkeeping system with symbols. One kind of symbol is the imaginary number [ square root (-1) ]. Imaginary numbers and complex numbers help physicists with understanding a lot of physical phenomena. As for your attempt to understand how 0.999... = 1 You made a good try. The numerals we use to represent numbers are the problem. In this case, remember that 1/3 = 0.333... and if you multiply both sides of that equation by 3 you will get 1 = 0.999... and there is your proof really simple ! "Fred" gave a very good description of a few of the characteristics of "infinity" One way to think of infinity is "increases without bound" - it's not a number, it's a description. When you are talking about the number line, it's where you want to go if you could. That will be how you interpret infinity when learning about the limit operator.