Congruence of Intergers and modular arthimetic

[doggie_Walkes]

Hey I am just wondeirng if I have to prove a congruence,

such as

c^3 is congruent to d modulo 7,

where d is set of {0,1 ,7}

So in this problem to prove this example all I need to do is prove that it is a equivalence relation?

So it is reflexsive, symmetric, and transitive?

Is this correct?

 

[Mark44]

Assuming c is an integer, I believe that what you're trying to prove is that
$$c^3 \equiv 0~mod~7$$
or
$$c^3 \equiv 1~mod~7$$

If so, it's not true. 3³ = 27 $\equiv$ 6 mod 7, and 5³ = 125$\equiv$ 6 mod 7.

 

[doggie_Walkes]

Thanks mark. I just had a bit of another question if I could ask you ask well?

It just how do i show

b^3 +b^2 +1 does not divide by 5

how do i prove it.

Im thinking this way,
cause i know that b^3 +b^2 +1 is not congruent to 0(mod5)

therefore we use contradition to prove it. I am just not sure how to use contradition? or maybe I am looking at this in a completely bad light? maybe there is another method?

 

[HallsofIvy]

If you meant "$c^3$ is congruent to one of 0, 1, 6 (mod 7)" then a perfectly valid way to do it is to look at all 7 possibilities: $0^3= 0$, $1^3= 1$, $2^3= 8= 1$, $3^3= 27= 6$, $4^3= 64= 1$, $5^3= 125= 6$, $6^3= 216= 6$, all "mod 7".

 

[Mark44]

If b^3 + b^2 + 1 is divisible by 5, the ones' digit in b^3 + b^2 + 1 has to be 0 or 5. Another way to say this is that b^3 + b^2 + 1 $\equiv$ 0 mod 5.

Work with the integers modulo 5.
If b $\equiv$ 0 mod 5, then b^3 $\equiv$ 0 mod 5, b^2 $\equiv$ 0 mod 5, so b^3 + b^2 + 1 $\equiv$ 1 mod 5. This means that the ones' digit has to be either 1 or 9.

If b $\equiv$ 1 mod 5, then b^3 $\equiv$ 1 mod 5, b^2 $\equiv$ 1 mod 5, so b^3 + b^2 + 1 $\equiv$ 3 mod 5. This means that the ones' digit has to be either 3 or 8. Because b^3 + b^2 + 1 is always odd, you'll never get an 8 digit in the ones' place, so for this case, the ones' digit has to be 3.

Continue this process for the other three equivalence classes to complete this proof.

 

[doggie_Walkes]

Hey mark, sorry for the late reply,

Im just wondering what you mean by ones' digit

 

[Mark44]

In the decimal number system, each digit in the numeric representation indicates a power of 10. For example, 435 = 4 * 10² + 3 * 10¹ + 5 * 10⁰. So 435 is 4 hundreds + 3 tens + 5 ones. The digit in the ones' place is 5 for this number.