Square root of 2 divided by 0 is rational?

In summary, Alan F. Beardon says that if ##a^2 - 2b^2 = 0## then ##\sqrt{2} = \frac{a}{b}## is rational.
  • #1
PcumP_Ravenclaw
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4
Dear All,
Please help me understand how ## \sqrt{2} ## divide by 0 is rational as stated in the excerpt from alan F beardon's book?
 

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  • #2
I don't follow his logic, but in any case it is actually undefined, and thus it is not meaningful to talk about it being rational or irrational. I think probably what he is saying is more complex than that "(sqrt 2) / 0 is rational"
 
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  • #3
He is saying that the multiplicative inverse of the element ##x=a+b\sqrt{2}## in the field ##\mathbb{Q}(\sqrt{2})## is the element ##x^{-1}=\frac{a}{a^2-2b^2}+\frac{-b}{a^2-2b^2}\sqrt{2}##.

There part where he mentions that ##a^2-2b^2\neq 0## is so that we can rest assured that the formula that he's given us for ##x^{-1}## is defined for all pairs of rationals ##a## and ##b##.
 
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  • #4
PcumP_Ravenclaw said:
Dear All,
Please help me understand how ## \sqrt{2} ## divide by 0 is rational as stated in the excerpt from alan F beardon's book?

You've misunderstood him. If ##a^2 - 2b^2 = 0## then ##\sqrt{2} = \frac{a}{b}## is rational. So, you know that ##a^2 - 2b^2 \ne 0##
 
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  • #5
Hey PeroK, I think this is what you mean??

##
a^2 - 2b^2 = 0
##
can be written as
##
(a + \sqrt{2}b)(a - \sqrt{2}b) = 0
##
here note that a and b are rational numbers ## {a + b√2 : a, b ∈ Q}, ##
so either ## (a + \sqrt{2}b)## or ## (a - \sqrt{2}b) ## must equal 0.
lets say ## (a - \sqrt{2}b) = 0 ## then ## \sqrt{2} = a/b ## which is a contradiction so ## (a - \sqrt{2}b) = 0 ## cannot happen.
 
  • #6
PcumP_Ravenclaw said:
Hey PeroK, I think this is what you mean??

##
a^2 - 2b^2 = 0
##
can be written as
##
(a + \sqrt{2}b)(a - \sqrt{2}b) = 0
##
here note that a and b are rational numbers ## {a + b√2 : a, b ∈ Q}, ##
so either ## (a + \sqrt{2}b)## or ## (a - \sqrt{2}b) ## must equal 0.
lets say ## (a - \sqrt{2}b) = 0 ## then ## \sqrt{2} = a/b ## which is a contradiction so ## (a - \sqrt{2}b) = 0 ## cannot happen.

That's not quite the way I'd do it! I'd say:

##a^2 - 2b^2 = 0 \ \ \Rightarrow \ \ a^2 = 2b^2 \ \Rightarrow \ \ \frac{a^2}{b^2} = 2 \ \Rightarrow \ \ (\frac{a}{b})^2 = 2##
 
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Related to Square root of 2 divided by 0 is rational?

1. What is the square root of 2 divided by 0?

The square root of 2 divided by 0 is undefined.

2. Why is the square root of 2 divided by 0 undefined?

This is because division by 0 is undefined in mathematics. It is considered to be a mathematical error and does not produce a meaningful result.

3. Can the square root of 2 divided by 0 be rational?

No, it cannot be rational because rational numbers are defined as numbers that can be expressed as a ratio of two integers, and dividing by 0 does not produce a rational number.

4. Is it possible to find a solution for the square root of 2 divided by 0?

No, there is no solution for this expression as division by 0 is undefined and does not have a meaningful answer.

5. How does dividing by 0 relate to the concept of infinity?

Dividing by 0 is often associated with infinity because as the divisor (in this case, 0) gets closer and closer to 0, the resulting quotient gets larger and larger, approaching infinity. However, this does not mean that dividing by 0 is equal to infinity, as it is still undefined.

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