How to Prove That the Square Root of Two Is Irrational
Assume that2{\displaystyle {\sqrt {2}}} is rational., Square both sides., Multiply both sides by b2{\displaystyle b^{2}}., Note that a2{\displaystyle a^{2}} is an even number., Substitute a=2k{\displaystyle a=2k} into the original equation., Expand...
Step-by-Step Guide
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Step 1: Assume that2{\displaystyle {\sqrt {2}}} is rational.
Then it can be expressed as a fraction ab{\displaystyle {\frac {a}{b}}}, where a{\displaystyle a} and b{\displaystyle b} are both whole numbers, and b{\displaystyle b} is not 0{\displaystyle 0}.
Furthermore, this fraction is written in simplest terms, meaning that either a{\displaystyle a} or b{\displaystyle b}, or both are odd whole numbers. 2=ab{\displaystyle {\sqrt {2}}={\frac {a}{b}}} -
Step 2: Square both sides.
2=a2b2{\displaystyle 2={\frac {a^{2}}{b^{2}}}} , 2b2=a2{\displaystyle 2b^{2}=a^{2}} , a2{\displaystyle a^{2}} is even number because it is equal to two times a whole number.
Since a2{\displaystyle a^{2}} is even, a{\displaystyle a} must be even too, because if it were odd, a2{\displaystyle a^{2}} would be odd as well (an odd number times and odd number is always an odd number). a{\displaystyle a} is even, so that means it can be written as two times a certain whole number, or in other words, a=2k{\displaystyle a=2k}, where k{\displaystyle k} is this whole number. , 2=(2k)2b2{\displaystyle 2={\frac {(2k)^{2}}{b^{2}}}}. , (2k)2=22k2=4k2{\displaystyle (2k)^{2}=2^{2}k^{2}=4k^{2}}. 2=4k2b2{\displaystyle 2={\frac {4k^{2}}{b^{2}}}} , 2b2=4k2{\displaystyle 2b^{2}=4k^{2}}. , b2=2k2{\displaystyle b^{2}=2k^{2}} , b2{\displaystyle b^{2}} is even number because it is equal to two times a whole number.
Since b2{\displaystyle b^{2}} is even, b{\displaystyle b} must be even too, because if it were odd, b2{\displaystyle b^{2}} would be odd as well (an odd number times and odd number is always an odd number). , You have just proven that b{\displaystyle b} is even.
However, you have also proven that a{\displaystyle a} is an even number.
This is a contradiction because in the beginning of this proof, it was assumed that ab{\displaystyle {\frac {a}{b}}} was written in simplest terms, but if both a{\displaystyle a} and b{\displaystyle b} are even, the numerator en denominator can be divided by 2, which means it was not written in simplest terms.
Since this is a contradiction, the original assumption that 2{\displaystyle {\sqrt {2}}} is rational is false, thus leading to the conclusion that 2{\displaystyle {\sqrt {2}}} is irrational. -
Step 3: Multiply both sides by b2{\displaystyle b^{2}}.
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Step 4: Note that a2{\displaystyle a^{2}} is an even number.
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Step 5: Substitute a=2k{\displaystyle a=2k} into the original equation.
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Step 6: Expand (2k)2{\displaystyle (2k)^{2}}.
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Step 7: Multiply both sides by b2{\displaystyle b^{2}}.
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Step 8: Divide both sides by two.
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Step 9: Note that b2{\displaystyle b^{2}} is an even number.
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Step 10: Recognise that this is a contradiction.
Detailed Guide
Then it can be expressed as a fraction ab{\displaystyle {\frac {a}{b}}}, where a{\displaystyle a} and b{\displaystyle b} are both whole numbers, and b{\displaystyle b} is not 0{\displaystyle 0}.
Furthermore, this fraction is written in simplest terms, meaning that either a{\displaystyle a} or b{\displaystyle b}, or both are odd whole numbers. 2=ab{\displaystyle {\sqrt {2}}={\frac {a}{b}}}
2=a2b2{\displaystyle 2={\frac {a^{2}}{b^{2}}}} , 2b2=a2{\displaystyle 2b^{2}=a^{2}} , a2{\displaystyle a^{2}} is even number because it is equal to two times a whole number.
Since a2{\displaystyle a^{2}} is even, a{\displaystyle a} must be even too, because if it were odd, a2{\displaystyle a^{2}} would be odd as well (an odd number times and odd number is always an odd number). a{\displaystyle a} is even, so that means it can be written as two times a certain whole number, or in other words, a=2k{\displaystyle a=2k}, where k{\displaystyle k} is this whole number. , 2=(2k)2b2{\displaystyle 2={\frac {(2k)^{2}}{b^{2}}}}. , (2k)2=22k2=4k2{\displaystyle (2k)^{2}=2^{2}k^{2}=4k^{2}}. 2=4k2b2{\displaystyle 2={\frac {4k^{2}}{b^{2}}}} , 2b2=4k2{\displaystyle 2b^{2}=4k^{2}}. , b2=2k2{\displaystyle b^{2}=2k^{2}} , b2{\displaystyle b^{2}} is even number because it is equal to two times a whole number.
Since b2{\displaystyle b^{2}} is even, b{\displaystyle b} must be even too, because if it were odd, b2{\displaystyle b^{2}} would be odd as well (an odd number times and odd number is always an odd number). , You have just proven that b{\displaystyle b} is even.
However, you have also proven that a{\displaystyle a} is an even number.
This is a contradiction because in the beginning of this proof, it was assumed that ab{\displaystyle {\frac {a}{b}}} was written in simplest terms, but if both a{\displaystyle a} and b{\displaystyle b} are even, the numerator en denominator can be divided by 2, which means it was not written in simplest terms.
Since this is a contradiction, the original assumption that 2{\displaystyle {\sqrt {2}}} is rational is false, thus leading to the conclusion that 2{\displaystyle {\sqrt {2}}} is irrational.
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