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Proof By Contradiction: There are no integer solutions to 14x+6y=1.

Here we will be walking through how to do a proof by contradiction. We will only be using definitions and properties of integers to show this.

Theorem

There are no integer solutions to the equation 14x+6y=1.

In the last two proofs that we did, we rewrote the theorem using symbolic logic, wrote down any needed definitions and made a table listing everything we were given and everything we needed to find. Here, we will try to work through this in one step since there is less going on than there was in the previous problems.

Proof Outline

We will be using a proof by contradiction, so this means that we will assume the negation of our theorem and show that this will lead to a contradiction. We, therefore, need the negation of the theorem. In this case, the negation of having no solutions would be that there does exist some solution. We now begin the proof with

  • Assume that there is exists an integer solution to 14x+6y=1.

This would then mean that there are two integers \(k_{1}\) and \(k_{2}\) such that \(14k_{1}+6k_{2}=1\). Because we are using a proof by contradiction, we don’t know exactly what we are trying to show, other than that we need to show something that can’t be true. In this case, we notice that if we divide both sides by 2, we get \(7k_{1}+3k_{2}=\frac{1}{2}\). Now, on the left hand side we have the product and sum of integers, so we have an integer. On the right side, we have \(\frac{1}{2}\), so if we combine this we are saying that \(\frac{1}{2} \in \mathbb{Z}\). However, this is clearly not the case, so we have our contradiction. We now need only write up our proof.

Proof

By way of contradiction assume that \(14x+6y=1\) has an integer solution. This means that there exists a \(k_{1},k_{2} \in \mathbb{Z}\) such that \(14k_{1}+6k_{2}=1\). However, if we divide both sides by 2, we get that \(7k_{1}+3k_{2}=\frac{1}{2}\). Note that \(7k_{1}+3k_{2}\) is the product and sum of integers, and is, therefore, an integer. Since this is also equal to \(\frac{1}{2}\), we also get the \(\frac{1}{2}\) is an integer. This is a contradiction since we know that \(\frac{1}{2}\) is not an integer. Hence \(14x+6y=1\) has no integer solutions.

Conclusion

Here, the statement was simpler to work with and negate, so our proof did go a little smoother than the last two. Even though we did shorten the process, we still followed the same outline of a proof we had in the previous cases. The exception here is that we showed that the negation of our original statement must be false, thus giving us that the original statement was true.

I hope that you learned something about the process of writing proofs and that this will help you in your studies. For more help visit here. If you enjoyed the post or learned something, make sure to like it and share with anyone else that may need help on similar topics.

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