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3 years ago

Drawing Conclusions

Mathematics

1 answer:

maksim [4K]3 years ago

3 0

Answer:

Some of the Exponents = -2 that is true, not 2.

Step-by-step explanation:

Let's check one at a time.

(a)The 6 without an exponent is equivalent to the 6 having a 0 exponent.

$6^{0} =1$ and $6^{1}$ = 6 (no exponent. 6 $\neq$ 1 therefore this statement is False.

(b)The sum of the exponents is -2.

let's check , if the base is same we can add the exponents that is the exponent rule.(well established).

if we add exponents in the given expression we get.

$6^{1} 6^{0} 6^{3} =6^{1+0+(-3)} =6^{-2}$ , therefore we can see that the sum of the exponents = -2 this is true.

(c) An equivalent expression is 65.6-7, lets evaluate our above expression, it is equal to $\frac{1}{36}$ which we can see that $\frac{1}{36} \neq 65.6-7$ ,therefore this statement is false as well.

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Answer:

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Step-by-step explanation:

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Proof by Contrapositive:

Assumed statement: Suppose that integers x and y do not have opposite parity.

Proven Statement: x and y are not a pair of consecutive integers.

Proof -

x = 2u₁ , y = 2u₂

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Not possible

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Proof by Contradiction:

Assumed statement: Suppose x and y are not a pair of consecutive integers.

Proven Statement: Suppose x and y do not have opposite parity.

Proof -

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⇒ either x and y are odd or even

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⇒x and y have same parity

Contradiction

If x and y are even

⇒x and y have same parity

Contradiction

(b)

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Assumed statement: Let n be an integer such that n is not odd (i.e. n is an even integer)

Proven Statement: n² is not odd (i.e n² is even)

Proof -

Let n is even

⇒n = 2m

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Assumed statement: Let n be an integer such that n² be odd.

Proven Statement: suppose that n is not odd (i.e n is even)

Proof -

Let n² is odd

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⇒2 | n²

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