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Question Number 36011 by solihin last updated on 27/May/18

$$\mathrm{simplify}:{}^{\prime }interval{number}^{\prime }$$$$(1,6)\cup (3,7)$$

Commented by Rasheed.Sindhi last updated on 27/May/18

$$\mathrm{For}\mathrm{overlapping}\mathrm{intervals}(a,b)\mathrm{\&}(c,d)$$$$(a,b)\cup (c,d)=(\min (a,c),\max (b,d))$$

Commented by Rasheed.Sindhi last updated on 27/May/18

$$(1,6)\cup (3,7)=(1,7)$$

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