atgnybo4fq

Answered

2022-11-22

Prove that square root of 6 is irrational by contradiction .

Answer & Explanation

Pignatpmv

Expert

2022-11-23Added 22 answers

Assume that $\sqrt{(}6)$ is rational.

Then $\sqrt{(}6)=\frac{p}{q}$ where p and q are coprime integers.

${\sqrt{6}}^{2}=6=\frac{{p}^{2}}{{q}^{2}}$

${p}^{2}=6{q}^{2}$

therefore ${p}^{2}$ is an even number since an even number multiplied by any other integer is also an even number. If ${p}^{2}$ is even then p must also be even since if p were odd, an odd number multiplied by an odd number would also be odd.

So we can replace p with 2k where k is an integer.

$(2k{)}^{2}=6{q}^{2}4{k}^{2}=6{q}^{2}2{k}^{2}=3{q}^{2}$

Now we see that $3{q}^{2}$ is even. For $3{q}^{2}$ to be even, ${q}^{2}$ must be even since 3 is odd and an odd times an even number is even. And by the same argument above, if ${q}^{2}$ is even then q is even.

So both p and q are even which means both are divisible by 2. But that means they are not coprime, contradicting our assumption so sqrt(6) is not rational.

Then $\sqrt{(}6)=\frac{p}{q}$ where p and q are coprime integers.

${\sqrt{6}}^{2}=6=\frac{{p}^{2}}{{q}^{2}}$

${p}^{2}=6{q}^{2}$

therefore ${p}^{2}$ is an even number since an even number multiplied by any other integer is also an even number. If ${p}^{2}$ is even then p must also be even since if p were odd, an odd number multiplied by an odd number would also be odd.

So we can replace p with 2k where k is an integer.

$(2k{)}^{2}=6{q}^{2}4{k}^{2}=6{q}^{2}2{k}^{2}=3{q}^{2}$

Now we see that $3{q}^{2}$ is even. For $3{q}^{2}$ to be even, ${q}^{2}$ must be even since 3 is odd and an odd times an even number is even. And by the same argument above, if ${q}^{2}$ is even then q is even.

So both p and q are even which means both are divisible by 2. But that means they are not coprime, contradicting our assumption so sqrt(6) is not rational.

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