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I have a question of how to solve this

$$B = \{x ∈ \mathbb{Z} : -2^x \leq 1\}$$

this according to a friend of mine should be $\{0, 1, 2, ... , \infty\}$

but how he gets to this is how I dont understand, if someone can help me understand this it would be great

Thank you, I dont know what the name of this is in English so I wrote something, feel free to edit

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    $(-2)^x$ or $-(2^x)$ ? – 2017-02-15
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    no paranthesis on the question – 2017-02-15
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    Thus, $-2^x$ is *negative*, because $2^x$ is always positive for $x \in \mathbb Z$. – 2017-02-15
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    Your friend is wrong: the answer is $\Bbb Z$, so: one $\infty$ too much and too few negative integers. – 2017-02-15
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    You friend would be right for $2^{-x}$. – 2017-02-15
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    thank you, but I dont quite understand how you know its Z, but Z is also negative numbers – 2017-02-15

1 Answers 1

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We have $-2^x \le 1$ iff $2^x \ge -1$

But always $2^x >0$, hence $B= \mathbb Z$

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    thanks, but then my friend is incorrect because he had a 0 in his answer? and second of all, do you confirm B=Z by checking the 2^x>0 or by testing x values? and if so, why is no negative numbers in the sequence? – 2017-02-15