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FUNDAMENTOS DE COMPUTACION

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Choose the main technique to prove the following statement.

If nnn is even, then 5n2+6n+75n2+6n+75n^2+6n+7 is odd.

T3: indirect proof by contradiction to show p≡Tp≡Tp \equiv T

T1: direct proof to show p→q≡Tp→q≡Tp \to q \equiv T

100%

T2: indirect proof by contraposition to show p→q≡Tp→q≡Tp \to q \equiv T

T4: two-step proof to show p↔q≡Tp↔q≡Tp \leftrightarrow q \equiv T

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Choose the main technique to prove the following statement.

If 5n2+6n+75n2+6n+75n^2+6n+7 is odd, then nnn is even.

T4: two-step proof to show p↔q≡Tp↔q≡Tp \leftrightarrow q \equiv T

T1: direct proof to show p→q≡Tp→q≡Tp \to q \equiv T

T2: indirect proof by contraposition to show p→q≡Tp→q≡Tp \to q \equiv T

T3: indirect proof by contradiction to show p≡Tp≡Tp \equiv T

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Consider a function f:R→Rf: R \to R, f(x)=x2f(x) = x^2.

Determine the correct sequence to prove the following statement.

ff is not injective.

If a statement is not used in the proof, you have to choose "Not used".

We cannot find any pre-image in the domain for f(x)=−1f(x)=-1.

Consider a real number −1-1 in the co-domain and let f(x)=−1f(x)=-1.

We can find two different pre-images in the domain: x=1x=1 and x=−1x=-1.

ff is not surjective

Consider a real number 1 in the co-domain and let f(x)=1f(x) = 1.

ff is not injective

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Consider a function f:R→Rf:R→Rf: R \to R, f(x)=2x+8f(x)=2x+8f(x) = 2x+8.

Determine the correct sequence to prove the following statement.

fff is surjective.

If a statement is not used in the proof, you have to choose "Not used".

a=ba=ba = b

x=y−82x=y−82x = \frac{y-8}{2} exists

Consider an arbitrary real number yyy.

Assume that f(a)=f(b)f(a)=f(b)f(a)=f(b).

Consider two arbitrary real numbers aaa and bbb.

fff is surjective

Assume f(x)=yf(x)=yf(x)=y.

2a+8=2b+82a+8=2b+8

2x+8=y2x+8=y

ff is injective

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Consider a function f:R→Rf: R \to R, f(x)=x2f(x) = x^2.

Determine the correct sequence to prove the following statement.

ff is not surjective.

If a statement is not used in the proof, you have to choose "Not used".

Consider a real number 1 in the co-domain and let f(x)=1f(x) = 1.

We cannot find any pre-image in the domain for f(x)=−1f(x)=-1.

Consider a real number −1-1 in the co-domain and let f(x)=−1f(x)=-1.

ff is not surjective

We can find two different pre-images in the domain: x=1x=1 and x=−1x=-1.

ff is not injective

View this question

Consider a function f:R→Rf: R \to R, f(x)=2x+8f(x) = 2x+8.

Determine the correct sequence to prove the following statement.

ff is injective.

If a statement is not used in the proof, you have to choose "Not used".

2a+8=2b+82a+8=2b+8

a=ba = b

Assume that f(a)=f(b)f(a)=f(b).

ff is injective

x=y−82x = \frac{y-8}{2} exists

2x+8=y2x+8=y

Assume f(x)=yf(x)=y.

Consider two arbitrary real numbers aa and bb.

ff is surjective

Consider an arbitrary real number yy.

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Find the correct answer for the following expression.

The ceiling of -6.8

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Find the correct answer for the following expression.

The ceiling of 1.6

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Find the correct answer for the following expression.

The floor of 8.3

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Find the correct answer for the following expression.

The floor of 5.4

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