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

#### SB257426

##### New Member
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))

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

Any help would be appreciated

jesus

#### tywebb

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

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

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

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

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

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

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

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

(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 $\bg_white n^2$, never 3 mod 4)

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

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

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

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#### SB257426

##### New Member
might be easier to use mod 4 arithmetic for $\bg_white n^2=4m+3$

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

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

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

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

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

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

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

(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 $\bg_white n^2$, never 3 mod 4)

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

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

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

##### Active Member
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

#### Pethmin

##### Well-Known Member
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

#### Pethmin

##### Well-Known Member
oh yeah myb I saw k on there and immediately thot induction

#### hscgirl

##### Active Member
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…

#### Pethmin

##### Well-Known Member
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

##### Active Member
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

##### Active Member
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

##### New Member
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

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