Square Floor!

Algebra Level 3

If s = ⌊ 2 − 3 ⌋ s = \lfloor \sqrt{2} - \sqrt{3} \rfloor and t = ⌊ 2 ⌋ − ⌊ 3 ⌋ t = \lfloor \sqrt{2} \rfloor - \lfloor \sqrt{3} \rfloor , which of the following statement is true?

Notation: ⌊ ⋅ ⌋ \lfloor \cdot \rfloor denotes the floor function .

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t < s t < s t > s t > s t = s t = s

Fidel Simanjuntak by Fidel Simanjuntak , 19, Indonesia

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1 solution

We can say that 2 ≈ 1.4 \sqrt{2} \approx 1.4 and 3 ≈ 1.7 \sqrt{3} \approx 1.7 .

Then, s = ⌊ 1.4 − 1.7 ⌋ → s = ⌊ − 0.3 ⌋ s = \lfloor 1.4 - 1.7 \rfloor \rightarrow s = \lfloor -0.3 \rfloor

s = − 1 s = -1 .

Otherwise, t = ⌊ 1.4 ⌋ − ⌊ 1.7 ⌋ → t = 1 − 1 = 0 t = \lfloor 1.4 \rfloor - \lfloor 1.7 \rfloor \rightarrow t = 1-1 =0 .

Hence, t > s t > s

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