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2. Construct derivations to prove that the following pair = of wffs in S are equivalent in SD: (20 ptstotal) A-(B ~ C) (A ^B)- C...

Question

2. Construct derivations to prove that the following pair = of wffs in S are equivalent in SD: (20 ptstotal) A-(B ~ C) (A ^B)- C

2. Construct derivations to prove that the following pair = of wffs in S are equivalent in SD: (20 ptstotal) A-(B ~ C) (A ^B)- C



Answers

Show that if $A$ and $B$ are similar, then $\operatorname{det} A=\operatorname{det} B$

For the first proof, we discussed them. Talk it out with you. And, um but you work out the details, but we have Zero is less than a is less than or equal to be. And we need to show that one over a It's greater than or equal to one over B. So, uh, first thing we can no is that A and B are both positive. Um, so that's important. And then, uh, just knowing that if you have a fraction or a reciprocal relationship, um, as a gets large, no one ever a get small. So the larger the denominator, the smaller the fraction. Okay, So because B is greater than or equal to a, that means it is a larger than or equal to denominator than a and so that would show that that would be true. Time for Part B. It's just the negative of that. So we have a is less than or equal to be, which is less than zero. That way, we want to show that one over a is greater than whatever be okay. So for negative numbers, um, as a gets large auras be gets large, one over a becomes a smaller fraction. But closer to zero is what you want to think about. So because a is ah, a larger negative number than be, um it's going to be closer to zero. Uh, which is what this means, like one over a would be closer to zero than one ever. Be okay. And now, part C. We know the A is less than zero is less than B. We need to really won over a 21 ever. Be. Yes, sir. For in this case, um, it's all about, like, the size of the because we don't know how far be is from zero, and we don't know how far away is from zero, but we do know that is negative. Okay, so and we do know that B is positive. So because a is negative and B is positive, we know that one over be, no matter how small of a fraction it ISS will always be greater than anything negative. Which is what one over a would be. Okay. So that's how those air related. Thank you very much.

Hello students. In this question we have to show that this A. C. It is nearly equal to this lambda, not bar divided by T. To S. And this holy scare. Okay so we can write that this A. C. It is equal to primary player by D. C. By two. And this holy square, okay? Or it can be written as by by four. And this D. C. This will be equals two lambda not bar divided by T. To S. And this holy square. And this tita by four. It is nearly equal to one. Okay. Which can be nearly taken as one. So this is the this will be nearly equal to the lambda, not bar divided by T. To S. And this holy square. So this becomes proof for this problem. Okay next we have to show that after solving it further this A. C. It is nearly equals to ellis square lambda note. Bar square developed A. S. Okay, so we can write that the A. S. Which is nearly equal to this dsc square and which can be taken as ellis square to square. So from here, after rearranging these two terms. So tita is can be taken from here and can be substituted here. So they see this will be equals two lamb not square. Let the so any square London, but not Bar square suburb. Yes. So this become proof for this problem. Okay. Thank you.

You know this problem? We've been given this similarly statement. Then we want to use it to complete this proportion here in the bottom. Now notice that B and E respond with G and a means you're on the bomber of G a notice that l corresponds with our. As I said, be corresponds with G. So be over G A. Because that'll be over RG.

This question asked us to utilize the periodic table and look at the periodic trend of atomic or ionic radius. One principle that's going to come in kind of handy here is that as we move top to bottom down any family on the periodic table, we tend to see an increase in radius whether it's for an eye on or for an atom. As long as we're comparing the same thing apples to apples, we see top to bottom down a family an increase in radius. So the first thing we want to apply those a little bit of a different rule. Want to look at why is it that the oxide and I on 02 minus is larger in size than oxygen and its atomic form, if we look at oxygen, it's element number eight here. So that means that has eight protons. That in its atomic form needs to hold eight electrons in its place in their place. Rather In the oxide, an ion, we have 10 electrons. Now, this is what I call the babysitting effect. Um the more kids you have, the harder it is to take care of them. In general, I should know. I'm from a family of nine kids and it, you know, it got a little tricky for my mom at times. So, um, if you have more kids, they're allowed to sort of wander off a little bit and it's more difficult for that nucleus in oxygen to hold a greater number of electrons close to itself in that wandering effect. His results in a larger size for an anti on versus the same species in its atomic form. This is always the case when you're comparing an anti onto the same adam, same number of protons means more electrons means larger size. And again, it's that babysitting effect that sort of kicks in here and leads to that principle. Um when we compare sulfide into oxide another in the same family and as we mentioned earlier, more towards the bottom of a family we go, the larger the radius is so for sulfur, whether it's in its anti on former its atomic form, when we're comparing it in this case to the oxide ion and the sulfite ion, sulfur is going to be bigger. It has more shells um than oxygen has. And those greater number of shells leads to a greater overall size for sulfur compared to oxygen or in this case for sulfide compared to oxide. How about when we compare sulfide to potassium? So now potassium has, we're told that the sulfide um an ion is larger in size than the potassium can I? And and why is that? They have the same number of electrons. They're both what we call Isil electronics, or have the same number of electrons in this case as are gone, noble gas. And the thing is who has the greater number of protons? The greatest number of protons between these two choices to hold that same number of electrons in place. potassium does potassium element number 19. While sulfur is element number 16, the greater number of protons to hold um the same number of electrons in place, potassium is going to be better at that. It's going to be able to hold those electrons more closely to itself, making a smaller size for potassium then for the anti on here in the sulphur form. And when we compare the potassium cat eye on to the calcium two plus cat eye on these are ISil electronic as well with argon. And who has the greater number of electrons protons rather to hold that same number of electrons in place. Let's just like see, calcium has one more proton element number 20 versus element number 19. And for the same reason potassium was able to hold those same number of electrons more closely to itself, resulting in a smaller size for potassium Catalan versus sulfide. The same thing here, the calcium to Pluskat ion has the same number of electrons, but more protons to hold that number of electrons in place, it's going to be more effective at it and we'll be able to hold those more closely toward its more abundant proton enriched nucleus than potassium is leading to a smaller size for the calcium to Pluskat ion than potassium plus. So, you see a general trend here is that the more negative the charge on an ion for Isil electronic species, same number of electrons. The larger that species becomes, the more positive the charge for can ions and ISil electronic species. The smaller um that species becomes


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