Hey Everyone! Is it correct to speak that the denominator of 0 is not defined, as 0 = 0/1 = 0/2 = 0/3.... Actually my teacher asked me the denominator of and I said that its not defined but she said it 1. Please reply soon and Thank You for answering in advance.

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TopNewestPlease review your posting. You first mention "denominator of 0 is not defined", but in your example you list it with 0 as the numerator, i.e. "\( 0 = 0/1 = 0/2 = 0/3 \ldots \)." Both of your statements are true, but express vastly different ideas.

I am also not certain what "but she said it 1" means. – Calvin Lin Staff · 3 years, 9 months ago

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He means that she ( his teacher) had said that the denominator should be 1. – Abhishek De · 3 years, 9 months ago

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Excellent question, Anoopam!

You might be thinking that the definition for rational number fails at the case of zero. Rational numbers are usually defined as \(\{ n\in \mathbb{Q}\iff n=\frac{p}{q}\) and \( (p,q)=1 \} \). So if \(0=\frac{0}{n}, n\ge 1\) then there's no problem as such.

But your definition does have some problems. If you see here you'll see the definition of \((0,n)\) is not universal. So the \((p,q)\) condition might fail.

Also the \((p,q)\) condition is imposed to to get unique representation for each rational number which fails in your case. In fact this 'unique' representation property is very useful to show bijections from rational to different sets.

If we make a more subjective treatment and see how we 'construct' rational numbers from positive integers, we'll see the usual bijection which is followed is \(f(1)=0, f(2)=1, f(3)=-1 , f(4)=-2, f(5)=-\frac{1}{2}...\). Thus \(0\) is in fact taken as \(\frac{0}{1}\). [See fig. for the bijection]

But in my opinion, there is no problem if we define zero to be vacuously a rational number and it doesn't 'need' to be represented as \(\frac{p}{q}\). – Abhishek De · 3 years, 9 months ago

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