Why is the intermediate value theorem a consequence of completeness and not just continuity?

Spivak saves the proof of the intermediate value theorem for his chapter on supremums and infemums.

His proof makes clear use of the completeness of reals, but why I struggle to see why it is dependent on completeness and not just continuity.

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On the rational numbers, the function 1×2>21_{x^2>2} is continuous.
– LutzL
2 days ago

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Continuity can be used in many sets that are not the reals
– Zelos Malum
2 days ago

  

 

@LutzL ah, yea there it is. I was assuming that we were always considering the reals. Thanks
– Demetri P
2 days ago

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1 Answer
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f(x)=x2−2
f(x) = x^2 – 2

This function is continuous at every rational number xx.
f(0)=−2andf(2)=+2.
f(0) = -2 \quad \text{and} \quad f(2) = +2.

If the intermediate value theorem could be proved using only continuity, then we would conclude there is some rational number x0x_0 between 00 and 22 for which f(x0)=0f(x_0)=0. But there is none.