Proof by Contradiction
Use Proof by Contradiction to show that there exists no integers x and y for which 6x+9y=1.
- There are integers x and y for which 6x+9y=1.
- If \(6x+9y=1\), then \(2x+3y=\frac{1}{3}\)
- This means that there cannot be integers x and y that meet this criteria as the sum of two integers will always be another integer.
Use Proof by Contradiction to show that there exists no integers \(x\) and \(y\) for which \(30x+20y=7\).
- There exists integers \(x\) and \(y\) for which \(10x+30y=7\)
- \(3x+2y=\frac{7}{10}\)
- This means that there cannot be integers x and y that meet this criteria as the sum of two integers will always be another integer.
Use Proof by Contradiction to show that \(\sqrt{ 3 }\) is irrational.
- \(\sqrt{ 3 }\) is rational, meaning it can be expressed in the form \(\frac{a}{b}\) where \(a\) and \(b\) are integers in their simplest form, with no common factors.
- \(3=\frac{a^2}{b^2}\)
\(3b^2=a^2\)
\(a^{2}\) is a multiple of 3, meaning \(a\) is a multiple of 3 too.
\(a=3k\)
\(3k^2=3b^2\)
\(9k^2=3b^2\)
\(b^2=3k^2\)
This means \(b^2\) is a multiple of 3, meaning \(b\) is a multiple of 3.
If \(a\) and \(b\) are both multiples of 3, this contradicts our original statement that \(a\) and \(b\) have no common factors.
Therefore \(\sqrt{ 3 }\) is irrational.
Use Proof by Contradiction to prove that the sum of a rational number and an irrational number is irrational.
- The sum of a rational number and an irrational number is rational.
- Let rational number = \(\frac{a}{b}\) , our irrational number = \(c\) where \(c \neq \mathbb{Q}\) and our sum = \(\frac{d}{e}\)
\(\frac{a}{b}+c=\frac{d}{e}\)
\(c=\frac{d}{e}-\frac{a}{b}\)
\(c=\frac{ab-ae}{eb}\)
This means \(c\) would be rational, which contradicts our original statement that \(c \neq \mathbb{Q}\)
Use Proof by Contradiction to show that \(\sqrt{ 2 }\) is irrational
- If \(\sqrt{ 2 }\) is rational then it can be expressed in the form \(\frac{a}{b}\) where a and b are integers
- So \(2=\frac{a^2}{b^2}\)
- So \(a^2=2b^2\)
- Since 2 is even, \(a^2\) must be even.
- This means that \(a\) must also be even.
- If a is even, then it can be expressed in the form \(2n\)
- So this can be written as \((2n)^2=2b^2\)
- \(4n^2=2b^2\)
- This means that \(b^2\) must be even, which means that \(b\) must also be even.
- If \(a\) and \(b\) are both divisible by 2, then they have a common factor of 2. This contradicts our original statement that \(a\) and \(b\) do not have any common factors and proves that \(\sqrt{ 2 }\) is irrational.