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명제와 조건

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Sundstrom, Mathematical Reasoning · 2.4절영어 원문

For each of the following statements

  • (1) Write the statement in the form of an English sentence that does not use the symbols for quantifiers. - (2) Write the negation of the statement in a symbolic form that does not use the negation symbol. - (3) Write the negation of the statement in the form of an English sentence that does not use the symbols for quantifiers.

(∀x∈R)(tan⁡2x+1=sec⁡2x)\left( \forall x \in \mathbb{R} \right) \left( \tan ^2 x + 1 = \sec ^2 x \right).

정답·풀이 보기 (영어 원문)
  • For each real number xx, tan⁡2x+1=sec⁡2x\tan ^2 x + 1 = \sec ^2 x. - (∃x∈R)(tan⁡2x+1≠sec⁡2x)\left( \exists x \in \mathbb{R}\right) \left( \tan ^2 x + 1 \ne \sec ^2 x \right). - There exists a real number xx such that tan⁡2x+1≠sec⁡2x\tan ^2 x + 1 \ne \sec ^2 x.

한 번 더 — 비슷한 문제Sundstrom, Mathematical Reasoning · 2.4절영어 원문

For each of the following statements

  • (1) Write the statement in the form of an English sentence that does not use the symbols for quantifiers. - (2) Write the negation of the statement in a symbolic form that does not use the negation symbol. - (3) Write the negation of the statement in the form of an English sentence that does not use the symbols for quantifiers.

(∀a∈R)(a+0=a)\left( \forall a \in \mathbb{R}\right) \left( a + 0 = a \right).

다 풀었으면 정답 확인
  • For each real number aa, a+0=aa + 0 = a. - (∃a∈R)(a+0≠a)(\exists a \in \mathbb{R})( a + 0 \ne a). - There exists a real number aa such that a+0≠aa + 0 \ne a.
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