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Problem (1), 10 points Determine if the following statements are True or False: You do not need to provide justification for YO1 answers on this problem:(a, 2 point...

Question

Problem (1), 10 points Determine if the following statements are True or False: You do not need to provide justification for YO1 answers on this problem:(a, 2 points) All subsets of uncountable sets are uncountable(b. 2 points) Bijective graphs cannot contain a 3-cycle2 points) Proving the inductive hypothesis is Tre is component of proof by induction(d, 2 points) If the range and co-domain of a function are not the same set, the function is not surjective2 points) If there are 14 people in cla

Problem (1), 10 points Determine if the following statements are True or False: You do not need to provide justification for YO1 answers on this problem: (a, 2 points) All subsets of uncountable sets are uncountable (b. 2 points) Bijective graphs cannot contain a 3-cycle 2 points) Proving the inductive hypothesis is Tre is component of proof by induction (d, 2 points) If the range and co-domain of a function are not the same set, the function is not surjective 2 points) If there are 14 people in class at least 3 people are guaranteed to have birthday in the exact same month



Answers

In Exercises 11 and $12,$ mark each statement True or False. Justify each answer.
a. The set of all affine combinations of points in a set $S$ is called the affine hull of $S$ .
b. If $\left\{\mathbf{b}_{1}, \ldots, \mathbf{b}_{k}\right\}$ is a linearly independent subset of $\mathbb{R}^{n}$ and if $\mathbf{p}$ is a linear combination of $\mathbf{b}_{1}, \ldots, \mathbf{b}_{k},$ then $\mathbf{p}$ is an affine combination of $\mathbf{b}_{1}, \ldots, \mathbf{b}_{k}$
c. The affine hull of two distinct points is called a line.
d. A flat is a subspace.
e. A plane in $\mathbb{R}^{3}$ is a hyperplane.

In this problem, we need to determine whether a given statement is true or false. Now given statement. And this problem is that the range of a function consists of just one number. That is domain also consists of just one number. Now, in order to determine whether or not this is true, let us consider a function F, which has a domain of our and a co domain of art. That is. This is a real valued function F on the set of real numbers. Now, let us define this function as F of X is equal to one for all X belongs to our. So that means that the function F is a constant function. Now we can see that the range of this function will consist of justice number one, because all of the numbers X in the domain are mapped to the number one. So the range is the singleton set consisting of only the number one, and thus the range consists of just one number. However, the domain is the set of real numbers, and this has infinitely many numbers in it. Thus, if the range of a function consists of just one number, then it is not necessary that the domain also consists of just one number. Thus the given statement is false, yeah.

In this problem, we need to determine whether the given statement is true or false. Now the given statement in this problem is that the composition of two art functions isn't even function now in order to determine whether or not this is true. Let us consider two functions F nd And let us consider that both of these functions are functions. Now we consider the composition of these two functions which is F compositional G. Now we need to determine whether or not this function is even now if composition G or minus X. If this is equal to have composition G of X, then have composition G will be an even function. So the value of this is equal to F O G o minus X. By using the definition of the composition of two functions. Now G is an art function, so the of minus X will be equal to minus G X. And thus this will become f minus G X, also F. S. And our function. So f minus G X will be equal to minus F of G of X. And if we use the definition of the composition of two functions, we can write this as minus F composition G of X. Hence we can see that if composition G of minus X is equal to minus F composition G of X, that means that if composition G is an art function, so it is not an even function, and hence the given statement is false.

In part first statement is given okay two or more three common for common five and three coma six or this joints there. So here you can see that three element or to coma three, bowman full cuomo life and three element oh said three comma six. So here you can see that said to coma three comma four to fly intersection three coma sixth equal to three. So it is not a this joint set. So here we can say it is a false statement now we sold four seconds said E E cuomo I oh come on you and E equal Moby we must see Kumar D all this joint said so here you can see that E element of E coma, e coma I coma oh coma you and E 11 oh E coma be comas T comedy e coma e coma I coma oh coma you intersection a coma be coma. See carmody equal to that A. So here you can see that it is not, it is joint safe so it is a false statement. It is our answer no. We sold part third that to coma six come on 10, come on 14 and three GoMA seven chroma 11 compound 15 all this joint say so here you can see that to coma six, coma 10 14 intersection said three cuomo seven former 11 go for 15 equal to say verify. So we can say this joint states too. It is it two statements discover answer no it's so part fools said to coma six cuomo 10 interesting three coma seven, coma 11 equal to why. So here we can say these are this joint said so it is a true statement. It is a quite a lot so

In this problem, we need to determine whether a given statement is true or false. Now in this problem, the given statement is that if the domain of a function consists of at least two numbers in the range also contains at least two numbers. Now, in order to determine whether or not this is true. Let us consider a function F. And let this function have a domain of E. And let it have a cool domain of art where the set E has only two points in it, say zero and one. So this function has a domain which has only two numbers. Now let us define this function as F of X is equal to 10 for all X belonging to eat. So we can see that F zero will be 10. NF one will also be equal to 10. Both numbers are mapped to the number 10, and that's the range Is the single concept consisting of only the number 10. So this function has a domain which consists of at least two numbers. In this case it contains exactly two numbers, but the range contains only one number. It does not contain at least two numbers. So that means it is possible for a function to have a domain with at least two numbers but arrange with less than two numbers. This means that the given statement is false.


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