Prove that √2 as a irrational number in a easiest way.
One of the easiest ways to prove the irrationality of √2 is the following proof by contradiction:
Assume √2 is rational:
√2 ∈ Q
Then we can write
√2 = p/q,
where p, q ∈ N (p and q are positive integers) and p/q is an irreducible fraction.
p and q then are coprime (en.wikipedia.org/wiki/Coprime_integers), i.e. their greatest common divisor (GCD) is 1.
Consequently the following holds
p/q = √2
(p/q)^2 = 2
(p^2)/(q^2) = 2/1
Given that p and q are coprime, p^2 and q^2 must also be coprime because the prime factorisation of p^2 contains nothing but the squares of the prime factors of p (or put another way two copies of every prime factor of p) and the same holds true for q^2 and q.
Thus (p^2)/(q^2) is an irreducible fraction. Observe that 2/1 is also an irreducible fraction since 1 is coprime with every integer (by definition).
Two irreducible fractions are equal if and only if both numerators and denominators are equal.
Thus p^2 = 2 and q^2 = 1.
Let S be the set of all square numbers (or perfect squares):
S = {n^2 | n ∈ N} = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, ...}
Given that p, q ∈ N it follows that p^2 ∈ S and q^2 ∈ S.
However p^2 = 2 and 2 is not a square number. Thus p^2 ∉ S.
p^2 ∉ {n^2 | n ∈ N}, where p ∈ N
That is a contradictory statement.
Thus the initial assumption that √2 is rational must have been false:
√2 ∉ Q
The square root of 2 is irrational.
Assume √2 is rational:
√2 ∈ Q
Then we can write
√2 = p/q,
where p, q ∈ N (p and q are positive integers) and p/q is an irreducible fraction.
p and q then are coprime (en.wikipedia.org/wiki/Coprime_integers), i.e. their greatest common divisor (GCD) is 1.
Consequently the following holds
p/q = √2
(p/q)^2 = 2
(p^2)/(q^2) = 2/1
Given that p and q are coprime, p^2 and q^2 must also be coprime because the prime factorisation of p^2 contains nothing but the squares of the prime factors of p (or put another way two copies of every prime factor of p) and the same holds true for q^2 and q.
Thus (p^2)/(q^2) is an irreducible fraction. Observe that 2/1 is also an irreducible fraction since 1 is coprime with every integer (by definition).
Two irreducible fractions are equal if and only if both numerators and denominators are equal.
Thus p^2 = 2 and q^2 = 1.
Let S be the set of all square numbers (or perfect squares):
S = {n^2 | n ∈ N} = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, ...}
Given that p, q ∈ N it follows that p^2 ∈ S and q^2 ∈ S.
However p^2 = 2 and 2 is not a square number. Thus p^2 ∉ S.
p^2 ∉ {n^2 | n ∈ N}, where p ∈ N
That is a contradictory statement.
Thus the initial assumption that √2 is rational must have been false:
√2 ∉ Q
The square root of 2 is irrational.
You just prove p and q are even, and contradict what you said in the whole beginning.
Another challenge is to prove π is irrational, because we can't use the same trick with even / even. It's not a piece of cake, or shall I say a piece of pie!
Another challenge is to prove π is irrational, because we can't use the same trick with even / even. It's not a piece of cake, or shall I say a piece of pie!
its the worst vpn.there LOL
i hate maths
@WassimBerbar said in #3:
> Another challenge is to prove π is irrational, because we can't use the same trick with even / even. It's not a piece of cake, or shall I say a piece of pie!
Au contraire, chēr barbare. One simply goes ontologique on this one. Dans nature, l'univers, il n'existe pas un seul circle. Hence, pi calculates the exact circumference of something that doesn't exist anywhere inside nature. That, monsieur, is as irrational as it gets!
> Another challenge is to prove π is irrational, because we can't use the same trick with even / even. It's not a piece of cake, or shall I say a piece of pie!
Au contraire, chēr barbare. One simply goes ontologique on this one. Dans nature, l'univers, il n'existe pas un seul circle. Hence, pi calculates the exact circumference of something that doesn't exist anywhere inside nature. That, monsieur, is as irrational as it gets!
@Megabarbaar said in #6:
> Au contraire, chēr barbare.
C'est un pun on ne peut plus mean.
> Dans nature, l'univers, il n'existe pas un seul circle. Hence, pi calculates the exact circumference of something that doesn't exist anywhere inside nature. That, monsieur, is as irrational as it gets!
Do you know what else is irrational? The absurdity of this thought process. If I use your philosophical approach, I can say "You can square a circle with compas and straightedge, because we can square something that doesn't exist in this physical world."
> Au contraire, chēr barbare.
C'est un pun on ne peut plus mean.
> Dans nature, l'univers, il n'existe pas un seul circle. Hence, pi calculates the exact circumference of something that doesn't exist anywhere inside nature. That, monsieur, is as irrational as it gets!
Do you know what else is irrational? The absurdity of this thought process. If I use your philosophical approach, I can say "You can square a circle with compas and straightedge, because we can square something that doesn't exist in this physical world."
@Thalassokrator said in #2:
> One of the easiest ways to prove the irrationality of √2 is the following proof by contradiction:
>
> Assume √2 is rational:
> √2 ∈ Q
>
> Then we can write
> √2 = p/q,
> where p, q ∈ N (p and q are positive integers) and p/q is an irreducible fraction.
> p and q then are coprime (en.wikipedia.org/wiki/Coprime_integers), i.e. their greatest common divisor (GCD) is 1.
>
> Consequently the following holds
> p/q = √2
> (p/q)^2 = 2
> (p^2)/(q^2) = 2/1
So far, so good. I would have used the same proposition. But the following is probably more complicated than it needs to be:
> Given that p and q are coprime, p^2 and q^2 must also be coprime because the prime factorisation of p^2 contains nothing but the squares of the prime factors of p (or put another way two copies of every prime factor of p) and the same holds true for q^2 and q.
>
> Thus (p^2)/(q^2) is an irreducible fraction. Observe that 2/1 is also an irreducible fraction since 1 is coprime with every integer (by definition).
> Two irreducible fractions are equal if and only if both numerators and denominators are equal.
>
> Thus p^2 = 2 and q^2 = 1.
>
> Let S be the set of all square numbers (or perfect squares):
> S = {n^2 | n ∈ N} = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, ...}
>
> Given that p, q ∈ N it follows that p^2 ∈ S and q^2 ∈ S.
>
> However p^2 = 2 and 2 is not a square number. Thus p^2 ∉ S.
>
> p^2 ∉ {n^2 | n ∈ N}, where p ∈ N
> That is a contradictory statement.
> Thus the initial assumption that √2 is rational must have been false:
>
> √2 ∉ Q
> The square root of 2 is irrational.
Of course this part is correct, but i think it can be done in an easier to understand way. Here it goes:
First, keep in mind that p and q are "coprime". If they aren't, we could reduce them to be so: if
√2 = x*p/x*q
we could cancel out the "x" and they are coprime.
If √2 = p/q then we can sqare both sides, getting
2 = p^2/q^2
we multiply both sides with q^2 and get
2*q^2 = p^2
From this follows that p^2 has to be an even number (it is 2 times something). Now, only even numbers sqared give even results (to get an even number by multiplication one of the factors has to be even. Since by "squaring" we multiply two identical factors both have to be even) and therefore p^2 has to be divisible by 4 and p has to be even. But if p is even and p^2 is divisible by 4 then we can divide the whole equation by 2 and get:
2*q^2 = p^2 = 4 * something
q^2 = 2 * something
and therefore q also has to be even, for the reasons stated above. That means, p and q are both divisible by 2 and therefore they are not coprime, as we have proposed at the beginning. They must have a common factor of 2. This is a contradiction and therefore p and q cannot exist, hence √2 is irrational.
(PS: That proof is actually the one Euklid came up with.)
> One of the easiest ways to prove the irrationality of √2 is the following proof by contradiction:
>
> Assume √2 is rational:
> √2 ∈ Q
>
> Then we can write
> √2 = p/q,
> where p, q ∈ N (p and q are positive integers) and p/q is an irreducible fraction.
> p and q then are coprime (en.wikipedia.org/wiki/Coprime_integers), i.e. their greatest common divisor (GCD) is 1.
>
> Consequently the following holds
> p/q = √2
> (p/q)^2 = 2
> (p^2)/(q^2) = 2/1
So far, so good. I would have used the same proposition. But the following is probably more complicated than it needs to be:
> Given that p and q are coprime, p^2 and q^2 must also be coprime because the prime factorisation of p^2 contains nothing but the squares of the prime factors of p (or put another way two copies of every prime factor of p) and the same holds true for q^2 and q.
>
> Thus (p^2)/(q^2) is an irreducible fraction. Observe that 2/1 is also an irreducible fraction since 1 is coprime with every integer (by definition).
> Two irreducible fractions are equal if and only if both numerators and denominators are equal.
>
> Thus p^2 = 2 and q^2 = 1.
>
> Let S be the set of all square numbers (or perfect squares):
> S = {n^2 | n ∈ N} = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, ...}
>
> Given that p, q ∈ N it follows that p^2 ∈ S and q^2 ∈ S.
>
> However p^2 = 2 and 2 is not a square number. Thus p^2 ∉ S.
>
> p^2 ∉ {n^2 | n ∈ N}, where p ∈ N
> That is a contradictory statement.
> Thus the initial assumption that √2 is rational must have been false:
>
> √2 ∉ Q
> The square root of 2 is irrational.
Of course this part is correct, but i think it can be done in an easier to understand way. Here it goes:
First, keep in mind that p and q are "coprime". If they aren't, we could reduce them to be so: if
√2 = x*p/x*q
we could cancel out the "x" and they are coprime.
If √2 = p/q then we can sqare both sides, getting
2 = p^2/q^2
we multiply both sides with q^2 and get
2*q^2 = p^2
From this follows that p^2 has to be an even number (it is 2 times something). Now, only even numbers sqared give even results (to get an even number by multiplication one of the factors has to be even. Since by "squaring" we multiply two identical factors both have to be even) and therefore p^2 has to be divisible by 4 and p has to be even. But if p is even and p^2 is divisible by 4 then we can divide the whole equation by 2 and get:
2*q^2 = p^2 = 4 * something
q^2 = 2 * something
and therefore q also has to be even, for the reasons stated above. That means, p and q are both divisible by 2 and therefore they are not coprime, as we have proposed at the beginning. They must have a common factor of 2. This is a contradiction and therefore p and q cannot exist, hence √2 is irrational.
(PS: That proof is actually the one Euklid came up with.)
@WassimBerbar Exactly.
@WassimBerbar
Á propos! Wassim!
If Mohammed can call Jesus a muslim,
I sure as hell can call you a barbarian!
Á propos! Wassim!
If Mohammed can call Jesus a muslim,
I sure as hell can call you a barbarian!
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