Question

Hello here is my tutorial on a proof :)

Answer 1

To proof: there are no natural numbers \(p^2\) & \(2q^2\) such that \(p^2=2q^2\)


Proof:
Case I: If p & q both are odd then
P & Q are odd =>\(p^2\) & \(q^2\) are odd
=>\(p^2\) is odd & \(2q^2\) is even
=>\(p^2\) & \(2q^2\) are not possible
Case II:if p is odd & q is even
since square of an odd natural number is odd & even natural number is even therefore
p is odd & q is even => \(p^2\) is odd & \(q^2\) is even
=> \(p^2\) is odd & \(2q^2\) is even
=> \(p^2\) & \(2q^2\) is not possible.

Answer 2

@waterineyes @mathslover @ParthKohli @angela210793 @lgbasallote @satellite73 @amistre64 @turingtest @ganeshie8 @Ganpat :)
Have a look :)

Answer 3

Nice proof.

Answer 4

thanx @ParthKohli :)

Answer 5

well it's also a little tutorial from me.
there are too many cases on this proof :)

Answer 6

@jiteshmeghwal9 if P^2 is even and q^2 is odd then 2q^2 will be even and p^2 will also be even . then in that case will u get any natural numbers as solution?

Answer 7

When p is even & q is odd:
In that case,
p is even=>p contains 2 as a factor
=>\(p^2\) contains \(2^2(=4)\) as a factor
q is odd=>\(q^2\)is odd
=>\(2q^2\) contains 2 as the only even factor
Hence,\(p^2\) contains 4 as a factor & \(2q^2\) contains 2 as the only even factor. hence, \(p^2\) & \(2q^2\) is not possible.

Answer 8

any suggestions will be appreciated :)

Answer 9

gt it @chandhuru ?

Answer 10

@jiteshmeghwal9 ya got it.. :)

Answer 11

I get the feeling that I'm going to learn more on this site than in my math class. Your tutorial looks awesome!

Answer 12

thanx @Crystalline :)

Answer 13

great @chandhuru :)

Answer 14

Great proof..

\[\huge \color{green}{Like..}\]

Answer 15

\[\Huge{\color{violet}{Me \space Too}}\]A medal\[\Huge{\color{gold}{\star}}\]

Answer 16

i like this tutorial

Answer 17

this can be also another method for showing that \(\sqrt{2}\) is irratioanal...

(with some little notations)

Answer 18

hmm...

Answer 19

What if both are even?

Answer 20

i love to know what is @jiteshmeghwal9 's answer to @sauravshakya 's question.

Answer 21

If q is even, then \( q = 2^n m \) and m is odd.
then
\[
2 q^2 = 2^{2n+1} m^2
\]
but \(2^{2n+1} m^2\) cannot be a square.

Answer 22

So q cannot be even. So this will cover the case where both are even.

Answer 23

\[2q^2=2^{2n+1}m^2 =(2^{n+\frac12}m)^2= p^2\]
\[p=2^{n+1/2}m\]
\(\therefore p\) is not a natural number

is this your argument @eliassaab ?

Answer 24

When P & Q both are even. then
P & Q both are even => P & Q both contains 2 as a factor
Let H.C.F of P & Q = D
Then, \(p^2=(dx)^2\) & \(2q^2=(2dy^2)\) for some natural numbers like 'x' & 'y' having no common factor other than 1.
Therefore, \(p^2=2q^2\)
\((dx)^2=2(dy)^2\)
\(d^2x^2=2d^2y^2\)
\(x^2=2y^2\)
since 'x' & 'y' have no common factor other than one so one of these natural number must be odd .
snce @ least one of them is odd. so reducing to one of the cases above we can say that \(p^2=2q^2\) is not possible.

Answer 25

@UnkleRhaukus @eliassaab @mukushla @sauravshakya :)

Answer 26

Thanx for loving to know my reply :)

Answer 27

\[\huge \text{Nice}\]

Answer 28

Thanx @mukushla :)

Answer 29

top stuff

Answer 30

thanx @UnkleRhaukus :)

Answer 31

this is a proof that root2 is irrational number

Answer 32

nope this proof helps to finding proof of \(\sqrt{2}\)

Answer 33

tht it is irrational

Answer 34

:) this same as p/q=root2, so there are no natural p and q that satisfy that.

Answer 35

yes!

Answer 36

NICE

Answer 37

thanx! @sauravshakya :)

Answer 38

@sasogeek have a look :)

Answer 39

It's a very good proof.
But it is not needed as for natural number
\[p ^{2} = q ^{2}\]
is true only when
\[p=q\]
Thus when you multiply square of p by two you do not obey the above rule and hence will never get any values for which the first equation is true.

Answer 40

\[1^2=(-1)^2\]
?

Answer 41

oh negative numbers are not natural numbers .

Answer 42

lol

Answer 43

HAHAHA :)