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Nataly_w [17]
3 years ago
11

We say that an integer a is a type 0 integer if there exists an integer n such that a = 3n. An integer a is a type 1 integer if

there exists an integer n such that a = 3n + 1. An integer a is a type 2 integer if there exists an integer n such that a = 3n + 2. Prove that if a is a type 1 integer, then a 2 is a type 1 integer\
Mathematics
2 answers:
Delicious77 [7]3 years ago
5 0

Answer:

<em>Proof below</em>

Step-by-step explanation:

Let's assume a is a type 1 integer. By definition, it means we can find an integer n such that

a=3n+1

We need to prove a^2 is a type 1 integer

Expanding

a^2=(3n+1)^2=9n^2+6n+1

If a^2 is a type 1 integer, then we should be able to find an integer m such as

a^2=3m+1

Equating

a^2=3m+1=9n^2+6n+1

solving for m

m=3n^2+2n

Since we know n is an integer, then the expression of m gives an integer also. Having found the required integer m, the assumption is proven

Gnoma [55]3 years ago
5 0

hope this helps :)

have an amazing day :)

Let's assume a is a type 1 integer. By definition, it means we can find an integer n such that

a=3n+1

We need to prove  is a type 1 integer

Expanding

If  is a type 1 integer, then we should be able to find an integer m such as

Equating

solving for m

Since we know n is an integer, then the expression of m gives an integer also. Having found the required integer m, the assumption is proven

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<em>
</em><em>Don't forget to rate this answer the Brainliest!</em>


7 0
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