I\'ve looked at the answer provided on the site and it\'s nothing close to what

Question

I've looked at the answer provided on the site and it's nothing close to what I came up with. I'm struggling in this class and I would appreciate it if someone could tell me where I went wrong with my solution. We were instructed in addition to the problem instructions that:

"Use only mod m, and congruence modulo m in your solution"

My solution:

Let p be the statement "a b (mod m)" and q be the statement "a mod m = b mod m" such that p q where m is a positive integer.

We will prove p q by contradiction.

Proof:

Assume p is true and ¬q is true.

Let a = 4, b = 2, m = 2.

Prove p: 4 2 ( mod 2 ) = 4 - 2 ( mod 2 ) = 2 mod 2 = 0. By definition this proves that a is congruent to b modulo 2.

Prove q (contradiction): 4 mod 2 = 2 mod 2

4 mod 2 = 0 and 2 mod 2 = 0 so therefore 0 = 0. This is true so the statement q is true while the statement p is true completing the contradiction.

My reasoning:

You only need to show one case where a contradiction exists as opposed to showing that every case is actually true. By the truth table of p q, if p is true and q is not true then the statement is false. Using proof by contradiction shows the statement is not false when p is true and q is not, leaving the only possible combinations of p and q that have the truth value of 1.

Explanation / Answer

PLEASE SEE MY COMMENTS IN CAPITALS ...

I've looked at the answer provided on the site and it's nothing close to what I came up with. I'm struggling in this class and I would appreciate it if someone could tell me where I went wrong with my solution. We were instructed in addition to the problem instructions that:...OK

"Use only mod m, and congruence modulo m in your solution"...THIS IS AN ORDER TO BE FOLLOWED!

My solution:

Let p be the statement "a b (mod m)" and q be the statement "a mod m = b mod m" such that p q where m is a positive integer.....OK

We will prove p q by contradiction.......OK

Proof:

Assume p is true and ¬q is true......OK

Let a = 4, b = 2, m = 2.......OK

Prove p: 4 2 ( mod 2 ) = 4 - 2 ( mod 2 ) = 2 mod 2 = 0. By definition this proves that a is congruent to b modulo 2......OK

Prove q (contradiction): 4 mod 2 = 2 mod 2... NO..

CHECK IF

4 MOD 2 IS NOT EQUAL TO 2 MOD 2 ..

4 mod 2 = 0 and 2 mod 2 = 0 so therefore 0 = 0. This is true so the statement q is true while the statement p is true completing the contradiction...

NO ..YOU BETTER WRITE ..

THE CONCLUSION WE GOT THAT 0 IS NOT EQUAL TO ZERO IS CONTRARY

TO BASIC FACTS..THIS HAS BEEN REACHED BY OUR ASSUMPTION OF

CONTRADICTION OF GIVEN STATEMENT ...SO OUR ASSUMPTION IS WRONG.

SO THE GIVEN STATEMENT IS TRUE..

My reasoning:

AT THE OUT SET WHY YOU ARE TALKING OF TRUTH TABLES, WHEN EXPRESSLY,

IT WAS STATED IN THE BEGINING THAT ...

"Use only mod m, and congruence modulo m in your solution"...THIS IS AN ORDER TO BE FOLLOWED!

OK..LET US TAKE IT THAT YOU WANT TO PROVE IT BY LOGIC..

You only need to show one case where a contradiction exists as opposed to showing that every case is actually true...........................OK

By the truth table of p q, if p is true and q is not true then the statement is false........OK

Using proof by contradiction shows the statement is not false when p is true and q is not, leaving the only possible combinations of p and q that have the truth value of 1......OK......

BUT PROOF BY TRUTH TABLE NEEDS ALL POSSIBLE COMBINATIONS OF P AND Q SHALL BE LISTED AND

IDENTICALITY HAS TO BE ESTABLISHED FOR DIFFERENT CONCLUSIONS IF THEY ARE TO BE PROVED AS SAME ..

WHAT DO YOU WANT PRECISELY?..IT IS NOT CLEAR

TELL US EXACTLY , WHAT IS BUGGING YOU SO THAT WE CAN CLARIFY IT ...

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