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Note: The Real numbers are an Archimedean field. At some point in your proofs the following basic property may be needed:Archimedean property: For every positive ...

Question

Note: The Real numbers are an Archimedean field. At some point in your proofs the following basic property may be needed:Archimedean property: For every positive real number a there exists a natural number such that n >

Note: The Real numbers are an Archimedean field. At some point in your proofs the following basic property may be needed: Archimedean property: For every positive real number a there exists a natural number such that n >



Answers

Prove Problems. True for all integers $n$ as specified. $-0<a<1$ implies that $0<a^{n}<1 ;$ for every natural number $n$

Want to show that this statement is true for all positive integers. So we'll use induction to do this. Ah, the base case would be on equals one, since it's a positive manager, then the left side would be equal to a 00 B and said the matrix with the one on the outside which we don't need to write on the right side, That would be a to the one which is just a and then be to the one which is just be so we can see that the left side and right side or the same so n equals one is true. We're gonna assume true, up to and equals kick. So we would write this as a 00 B to the power of K is equal to a to the K 00 B to the king, Then for n equals K plus one, we would have a 00 b to the power of K plus one and ah, based on exponents, we know that this is the same as a 00 p to the power of K times a 00 B and we can see that, uh, we could make use of our previous step. Now, this is the same as a to the K 00 b to the K times. Ah, a zero Ciro And be so for you to do the next step. You need to know your matrix Multiplication. So, uh, your first row is being multiplied by the first element in the second matrix. So it'll be written as a to the K Time's a plus zero times zero zero, and then you'll have the first row A to the K times second element in the second matrix, plus zero times zero and then ah, on the bottom, you've got the second row times the first element in the second matrix to zero time zero plus B to the K time zero and also zero times uh, B was b to the K times and you can see that this simplifies to a to the K plus one on the top, right will have zero bottom left will have zero bottom right will have b to the K plus one. And that's what we wanted to show eso. Therefore, we can conclude that a zero zero b to the power of N is equal to a to the n zero and zero B to the end for all and in the positive images

In this question, we are asked to prove using induction that the statement here is if we have a metric aye Looking like this is 00 b. Then a power bi end will be dismantled like we just power is entry by in when in is positive integer Um so first we look at the basic step is very easy when in equals, who won is just itself, right? So this is obviously true for inactives that we assume that the statement is true for some in so that formalized through for foreign. We want to show that the formalized true for the next situation for the plus one. So we just started. This is p of N we have this ass and assume fact. Then we multiply a bull's eye, right? It's a logical thing to do and you can see that the right hand side will become the form that be one. So this is actually the statement for in plus one. And so we have clear the inductive state as well. So with this to come by, we have proved the statement using math induction and that is it is very simple. Thank you

For the following problem, we're going to let X be greater than zero and then be greater than one. The real numbers. We want to prove that 1/1 plus X to the end. Okay. Is greater than one plus NX. Yeah. So we notice that let's take the this is when we have all values. So let's consider the derivative. We're going to get 10 times one class axe Ready to the Power of N -1. Yeah. And we know that N has to be greater than one. So we know that in any case this right here is always going to be um this is going to be a positive number. What we end up seeing here, it's not going to be a fraction. So we see in this case it will be greater than this and um that's going to be the final result because we see that this is going to grow faster since N is equal to one. We're getting up having this grow faster than this because this is linear and this is exponential. So it's our final answer.

We know that if we're given y equals the N through it of acts, then we know the domain. The function is the set of all real numbers. If an Izod when it is odd and exes, any real number of the range is also the set of all real numbers because we're looking at and equals two times some constant K plus one. Therefore, given this we know, the end is an odd number.


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