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This is a discussion on INTP and INTJ females unite! within the NT's Temperament Forum- The Intellects forums, part of the Keirsey Temperament Forums category; Originally Posted by FlaviaGemina OK, girls, here's another random question. Out of the following, who do you get along best ...
Online, I've enjoyed conversations with all four types. The only times I've felt dissonance are with individuals who are trying to fit into a stereotype that is clearly not close enough to their natural personalities to be comfortable. I'd say that's the case regardless of type.
I'm pretty terrible at typing people. I suspect one of my male friends is an INTJ and we get along well but I don't know if I know any of the other INTP/J types so can't say how I'd get along with the others.
@RedX, or anyone who's good at maths.
I need help. I'm pre-learning some maths for next school year to help my pupil.
Today I did about proofs.
Here's one of the questions and my answer. Can you tell me whether I did it right? The book doesn't give the answers for this chapter.
"Prove by direct argument that the sum of the squares of any 5 consecutive integers is divisible by 5".
m2 + (m+1)2 + (m+2)2 + (m+3)2 +(m+4)2 = x
m2 + ( m2+2m + 1) + ( m2+4m + 4) + ( m2+6m + 9) + ( m2+8m + 16) = x
5m2 + 20m + 30 = x
m2 + 4m + 6 = x/5
Then I put some random numbers for m to test it
e.g. m = 2
5*22 + 20*2 +30 = 20 + 40 + 30 = 90
22 + 4*2 + 6 = 4 + 8 + 6 = 18
90/5 = 18
Is this right? The book didn't really explain what "by direct argument" means. Some of the examples of it look more like "by exhaustion".
(... I'm also an NF ... I iz in your bases, NTs (Please don't hurt me))
It looks like you did it right, but I'm going to redo it just to check the arithmetic
m2 + (m+1)2 + (m+2)2 + (m+3)2 +(m+4)2 = q5 + r .... = r (mod 5) ... note: by the division algorithm q and r are unique (0<=r<5)
m2 + (m+1)2 + (m+2)2 + (m+3)2 +(m+4)2 = m2 + ( m2+2m + 1) + ( m2+4m + 4) + ( m2+6m + 9) + ( m2+8m + 16)
= 5m2 + 20m + 30 = (m2 + 4m + 6)5 = q*5 + r
r=0, q=m2 + 4m + 6 ... since m is an integer, q is an integer
i.e. There exists an integer q such that m2 + (m+1)2 + (m+2)2 + (m+3)2 +(m+4)2 = q*5
Yup ... you did it correctly ... directly just means that you start with the premise (The addition on 5 consecutive numbers is equal to an integer, c) and work your way to the conclusion (c is divisible by 5)
Another common proof technique is a subset of indirect proofs:
Assume c is not divisible by 5 ... now prove that assumption leads to a contradiction
By additions did you mean the division algorithm and/or modulo a number (mod 5)?
They are just other ways to look at whether or not a divides b
Divisor - Wikipedia, the free encyclopedia
Division Algorithm (link from wikipedia that can be used for two negative numbers too)
Euclidean division - Wikipedia, the free encyclopedia
Modular arithmetic - Wikipedia, the free encyclopedia
You don't need to write it down, but some people might be able to picture it easier by looking at the problem through the lens of these mathematical ideas ^__^
@Mr. Meepers , thanks. I don't think my kid will have to answer these questions to such a high level. (Or at least I hope he won't, because if I have to help him with it, that will create more work for me :) ) It's 'only' A-level (Year 13 at high school).