Refer to the following conditional statement: If a and b are odd integers, then a – b is an even integer. Identify the hypothesis. A. a – b is not an even integer B. a and b are not odd integers C. a and b are odd integers D. a – b is an even integer
I think its C
@whpalmer4
Yes that is correct
Ok, thanks
Refer to the following conditional statement: If a and b are odd integers, then a • b is an odd integer. Which is the inverse of the statement? A. a • b is an odd integer B. If a • b is an odd integer, then a and b are odd integers. C. If a and b are not odd integers, then a • b is not an odd integer. D. a and b are odd integers I think it's C.
@BlackLabel
Yup that is correct
If p then q Inverse: If ~p then ~q
Oh, then B? O.O
Wait, thats not right
Ya, its C
Right?
Yup
You got it right I was just giving you the definition of an inverse statement
Ok, thanks! Can you help with a few more?
Ya np. Just working at the same time so it may take a lil longer to respond
Ok, thanks! Refer to the following conditional statement: If a and b are odd integers, then a • b is an odd integer. Which is the contrapositive of the statement? A. If a • b is not an odd integer, then a and b are not odd integers. B. If a • b is an odd integer, then a and b are odd integers. C. a • b is an odd integer D. a and b are odd integers
I would say B
Oh wait.. actually, A
Yup it is A
Sorry! XP
What kind of reasoning is Kim using? Kim notices that 24 = 6 + 18, 63 = 9 + 54, 39 = 12 + 27, and 72 = 9 + 63, so she thinks it is likely that she can write every number divisible by 3 as the sum of two numbers divisible by 3. A. deductive B. inductive C. hypothesis D.conclusion
B?
Yup I guess we can say inductive
That is the most apropo even though we usually can not induce like that when proving
Ok, thanks! What kind of reasoning is used here? If a = b and c = d, then a • c = b • d. A. hypothesis B. conclusion C. deductive D. inductive I think this is C
@BlackLabel
I think so too since this is more general
Yup, thanks! How about this one? Which is a counterexample that disproves the following statement? For all integers x, 1/x \[\le\] x A. x = 5 B. x = 100 C. x = 2 D. x = 0.5
Oh, my bad, the: \[\le\] Should be after 1/x and before x, like this: 1/x <_ x
I'm not sure about this question.. can you explain?
Maybe D..?
Ok basically the statement is saying that for any integers x then \( \frac{1}{x}\) will always be smaller than or equal to x Now we need to find an example where this is not the case BUt the statement is true for 100 since 1/100<100 and 1/5<5 and 2< 1/2 Now the issue with .5 is that .5 is not an integer. .5 is a fraction and the question specifically states that x must be ab integer
1/.5=2 so .5<1/.5 But .5 is not an integer So D would be my preferred option eventhough its not really true
Ah, I see! Then D it is?
It really isnt D but just go for it anyways
Btw, did you realize that the question says: Which is a counterexample that DISPROVES the following statement?
Do, if you say it isn't D, then for this question, it IS D?
Yaaaa I got that None of them disprove the statement In order to disprove the statement The hypothesis must be true which is that x must be an integer. Since x is not an integer then x can not qualify to disprove the statement
I mean So*
Then its still D?
Lets go for D. They may have worded the question in correctly and meant to say ration number instead of integer
rational number*
Alright then! Thanks What is the reason for Statement 2? A. addition property of equality B. property of opposites C. identity property of addition D. associative property of addition
Is it B?
addition property of opposites The property that states the sum of a number and its opposite is always zero.
Yup that is what I was thinking
Only one that makes most sense
Ok, thanks :) This one: Write the contrapositive of the conditional statement and then determine if the conditional statement and the contrapositive are true or false. If 6x ≠ 24, then x ≠ 4. A. If 6x = 24, then x = 4. The conditional statement and the contrapositive are both true. B. If x = 4, then 6x = 24. The conditional statement and the contrapositive are both true. C. If x = 4, then 6x = 24. The conditional statement is false, and the contrapositive is true. D. If x ≠ 4, then 6x ≠ 24. The conditional statement and the contrapositive are both false.
Oh, sorry, its all mixed up O.O A. If 6x = 24, then x = 4. The conditional statement and the contrapositive are both true. B. If x = 4, then 6x = 24. The conditional statement and the contrapositive are both true. C. If x = 4, then 6x = 24. The conditional statement is false, and the contrapositive is true. D. If x ≠ 4, then 6x ≠ 24. The conditional statement and the contrapositive are both false.
I think B
My bad, I meant A
Im not sure Im flirting btwn A and B Lemme think abt this
Ok
"A" definitely sounds more correct. So I would go with A but maybe post another question and get someone to take a look at this
Aren't A and B almost the same thing? I mean, all they did was switch these around: If 6x = 24, then x = 4.
Ok, thanks :) I'll do that!
I would like to disagree the if part is the hypothesis: "A very closely related logical meaning is that for any statement or claim of the form "if A, then B", A is said to the be hypothesis of the claim. This usage transfers the idea of a "hypothesis" from the argument that establishes A⇒B to the naked assertion that A⇒B. "Assumption" is possible here, but appears to be less common than "hypothesis", especially if the claim is written symbolically rather than a theorem statement in prose."
You're right, that's the formal logic usage, the hypothesis being the antecedent. I withdraw my objection, though I'll continue to think that's a ridiculous hypothesis :-)
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