Can anyone please provide some feedback on my proof (1 Viewer)

SB257426 · Mar 19, 2023

The question was asking: Prove the following statement using either direct or contrapositive proof: If n is an integer then 4 does not divide n^2-3

Here is my working out:

let n^2 - 3 = 4m
By way of contradiction assume n^2 - 3 is rational, ie; n^2 - 3 = a/b (BTW in the funky looking expression for n the square root sign is not supposed to be over the 1/sqrt(p))

Screen Shot 2023-03-19 at 5.53.42 pm.jpg

Once again by way of contradiction, assume sqrt(p) is rational:

Screen Shot 2023-03-19 at 6.00.47 pm.jpg

Any help would be appreciated

SB257426 · Mar 19, 2023

inigomcvinish said:
jesus

Is it a bad proof lol

tywebb · Mar 19, 2023

might be easier to use mod 4 arithmetic for $n^2=4m+3$

If 4 divides $n^2-3$ then there exists an integer $m$ such that $n^2-3=4m\therefore n^2=4m+3$

$4m+3\equiv3(\mod 4)$

If $n$ is an integer then there are 4 cases for $n^2$

$n^2\equiv(0(\mod 4))^2=0(\mod 4)$

$n^2\equiv(1(\mod 4))^2=1(\mod 4)$

$n^2\equiv(2(\mod 4))^2=0(\mod 4)$

$n^2\equiv(3(\mod 4))^2=1(\mod 4)$

(or you could simplify this a bit just by looking at odd and even cases for n - and you only get 0 or 1 mod 4 for $n^2$ , never 3 mod 4)

In no case is $n^2$ going to be $3(\mod 4)$

Hence if 4 divides $n^2-3$ then $n$ is not an integer.

Contrapositive: If $n$ is an integer then 4 does not divide $n^2-3$

SB257426 · Mar 19, 2023

tywebb said:
might be easier to use mod 4 arithmetic for $n^2=4m+3$

If 4 divides $n^2-3$ then there exists an integer $m$ such that $n^2-3=4m\therefore n^2=4m+3$

$4m+3\equiv3(\mod 4)$

If $n$ is an integer then there are 4 cases for $n^2$

$n^2\equiv(0(\mod 4))^2=0(\mod 4)$

$n^2\equiv(1(\mod 4))^2=1(\mod 4)$

$n^2\equiv(2(\mod 4))^2=0(\mod 4)$

$n^2\equiv(3(\mod 4))^2=1(\mod 4)$

(or you could simplify this a bit just by looking at odd and even cases for n - and you only get 0 or 1 mod 4 for $n^2$ , never 3 mod 4)

in no case is $n^2$ going to be $3(\mod 4)$

Hence if 4 divides $n^2-3$ then $n$ is not an integer.

Contrapositive: If $n$ is an integer then 4 does not divide $n^2-3$

Oh lol I should have thought of that instead.........

Is there anything wrong with my proof though?, would I still full marks for such question (it was worth 3 marks)

s97127 · Mar 19, 2023

i think the easiest is as follows:

Assume n^2-3 = 4m
Case 1: n is even number, n = 2k
4k^2 - 3 = 4m is wrong because even number - odd number cannot be an even number
Case 2: n is an odd number, n = 2k+1
2k^2 + 2k - 1 = 2m is wrong because of the same reason.
Therefore n^2 - 3 is not divisible by 4

Average Boreduser · Mar 20, 2023

s97127 said:
i think the easiest is as follows:

Assume n^2-3 = 4m
Case 1: n is even number, n = 2k
4k^2 - 3 = 4m is wrong because even number - odd number cannot be an even number
Case 2: n is an odd number, n = 2k+1
2k^2 + 2k - 1 = 2m is wrong because of the same reason.
Therefore n^2 - 3 is not divisible by 4

thats induction tho n

s97127 · Mar 20, 2023

Pethmin said:
thats induction tho n

No it's not. Its a proof by contradiction

Average Boreduser · Mar 20, 2023

oh yeah myb I saw k on there and immediately thot induction

hscgirl · Mar 20, 2023

s97127 said:
No it's not. Its a proof by contradiction

the fact a y9 accelerated kid (or whatever he is lol) is smarter than a y10 accelerated kid…

Average Boreduser · Mar 20, 2023

hscgirl said:
the fact a y9 accelerated kid (or whatever he is lol) is smarter than a y10 accelerated kid…

bruh the dudes already finished school, he capping hard.

hscgirl · Mar 20, 2023

Pethmin said:
bruh the dudes already finished school, he capping hard.

idk what to believe on the internet anymore

yeah but that's true, im so confused why he doesnt at least change his hsc grad year from 2020 to N/A… (not to mention that u pethmin apparently graduated from the hsc in 1998

)

s97127 · Mar 20, 2023

hscgirl said:
idk what to believe on the internet anymore
yeah but that's true, im so confused why he doesnt at least change his hsc grad year from 2020 to N/A… (not to mention that u pethmin apparently graduated from the hsc in 1998 )

either way i'm still smarter than him lol

Sam14113 · Mar 22, 2023

SB257426 said:
The question was asking: Prove the following statement using either direct or contrapositive proof: If n is an integer then 4 does not divide n^2-3

Here is my working out:

let n^2 - 3 = 4m
By way of contradiction assume n^2 - 3 is rational, ie; n^2 - 3 = a/b (BTW in the funky looking expression for n the square root sign is not supposed to be over the 1/sqrt(p))

View attachment 38010
Once again by way of contradiction, assume sqrt(p) is rational:

View attachment 38011

Any help would be appreciated

Dunno if you still need help on this … if you don’t just ignore but otherwise here’s what I have to say

You said: By way of contradiction assume n^2-3 is rational.

The problem here is that the question isn’t asking us to prove it’s irrational. It’s asking us to prove it’s a multiple of 4. So if we want to contradict that we need to say, “Assume it’s n a multiple of 4”, rather than “assume it’s rational.”

the fact is, n^2-3 is in fact rational. In fact it’s integral.

this is a fairly easy mistake to make - you see lots of ‘irrationality’ proofs by contradiction in the 4U course, so that’s often where your brain goes when you see proof by contradiction. Remember though, the contradiction should directly contradict what you’re trying to prove.

hope that helps

zaccoo · Dec 1, 2023

s97127 said:
i think the easiest is as follows:

Assume n^2-3 = 4m
Case 1: n is even number, n = 2k
4k^2 - 3 = 4m is wrong because even number - odd number cannot be an even number
Case 2: n is an odd number, n = 2k+1
2k^2 + 2k - 1 = 2m is wrong because of the same reason.
Therefore n^2 - 3 is not divisible by 4

Your working is flawed on the 5th line. 2k^2 + 2k - 1 = 2m

s97127 · Dec 2, 2023

zaccoo said:
Your working is flawed on the 5th line. 2k^2 + 2k - 1 = 2m

Why? LHS is an odd number while RHS is even.

zaccoo · Dec 2, 2023

s97127 said:
Why? LHS is an odd number while RHS is even.

as in you can't get to that line, your working is wrong. if you write it out again and simplify you'll see what I mean

s97127 · Dec 3, 2023

zaccoo said:
as in you can't get to that line, your working is wrong. if you write it out again and simplify you'll see what I mean

Just did it again and still got that

kendricklamarlover101 · Dec 3, 2023

s97127 said:
Just did it again and still got that

they are probably referring to $(2k+1)^{2} = 2k^2 +2k +1$ which should be $(2k+1)^{2} = 4k^2 +4k +1$ . the proof is still correct otherwise

s97127 · Dec 3, 2023

kendricklamarlover101 said:
they are probably referring to $(2k+1)^{2} = 2k^2 +2k +1$ which should be $(2k+1)^{2} = 4k^2 +4k +1$ . the proof is still correct otherwise

N^2 - 3 = 4m where N = 2k+1
4k^2 + 4k + 1 - 3 = 4m
2k^2 +2k -1 = 2m

Can anyone please provide some feedback on my proof (1 Viewer)

SB257426

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