Root 4 is irrational!

Let us prove that 4 \sqrt 4 is irrational:

Let us assume that 4 \sqrt 4 is rational, then it can be expressed in the form of p q \frac pq where p , q p,q have no factors in common, thus p q = 4 \frac pq = \sqrt 4 . or p 2 / q 2 = 4 \color{#D61F06}{p^2 /q^2 = 4} then p 2 = 4 q 2 p^2 = 4q^2 , with p has a factor of 4 \color{#3D99F6} {\text{with } p \text{ has a factor of 4}} . Let p = 4 m p = 4m , then ( 4 m ) 2 = 4 q 2 16 m 2 = 4 q 2 , 4 m 2 = q 2 \color{#20A900}{(4m)^2 = 4q^2 } \rightarrow 16m^2 = 4q^2, 4m^2 = q^2 , then q also has a factor of 4 \color{#EC7300}{q \text{ also has a factor of 4}} . But this contradicts the fact that p , q p,q are coprimes. Thus the assumption that 4 \sqrt 4 is rational is wrong. Hence, 4 \sqrt 4 is also irrational.

This concludes that 4 = 2 \sqrt 4 = 2 is an irrational number.

Which of these above colored equations are not necessarily correct?

"1" represents r e d \color{#D61F06}{red} , "2" represents b l u e \color{#3D99F6}{blue} , "3" represents g r e e n \color{#20A900}{green} , "4" represents o r a n g e \color{#EC7300}{orange} .

2 and 4 2 and 3 1 and 3 1 and 4 3 and 4 1 and 2

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2 solutions

If a prime p p divides m 2 m^{2} , then p p also divides m m . Here, 4 4 is not a prime. That is the fallacy. Hence the statements that say that p p has a factor of 4 4 (i.e. statement (2) ) and q q has a factor of 4 4 (i.e. statement (4) ) are incorrect.

Moderator note:

How would you characterize the set of numbers p p such that:

If p p divides m 2 m^2 , then p p divides m m .

What numbers, other than the prime numbers, satisfies this property?

How do we show that 4 doesn't satisfy this property, (and hence "that is the fallacy").

I'm confused, I thought that p and q need not have a factor of 4, because √4=2 is enough to make p² and q² have a '4' in them. Isn't this enough to solve the problem?

karun mathews - 3 years, 3 months ago

Being colorblind this question is not answerable. I don't even see a red equation.

Sheldon Jolson - 3 years ago

Just take p to be 6 and q to be 3.

36 has a factor 4 but 9 doesn't.

I forgot to add that this problem is based on the fact that "If p-squared is divisible by a number A, then p is also divisible by A" is only valid if A is a prime number. So, the same proof would be correct for root-3, but not root-25.

Kamala Ramakrishnan - 6 years ago

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