Proof by Contradiction

This is example demonstrates how to write a proof by contradiction in MathML.

ContradictionProof.html

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		<p>Prove that there are an infinite number of prime numbers.</p>
		<p>To begin, we assume that there are only a finite number of prime numbers, say 
			<math><msub><mi>p</mi><mn>1</mn></msub><mo>,</mo>
						<msub><mi>p</mi><mn>2</mn></msub><mo>,</mo>
						<msub><mi>p</mi><mn>3</mn></msub><mo>,</mo>
						<mi>&hellip;</mi><mo>,</mo>
						<msub><mi>p</mi><mrow><mi>n</mi><mo>-</mo><mn>1</mn></mrow></msub><mo>,</mo>
						<msub><mi>p</mi><mi>n</mi></msub></math>.</p>
		<p>Consider the number <math><mi>q</mi><mo>=</mo>
			<msub><mi>p</mi><mn>1</mn></msub><mo>&times;</mo>
			<msub><mi>p</mi><mn>2</mn></msub><mo>&times;</mo>
			<msub><mi>p</mi><mn>3</mn></msub><mo>&times;</mo>
			<mi>&hellip;</mi><mo>&times;</mo>
			<msub><mi>p</mi><mrow><mi>n</mi><mo>-</mo><mn>1</mn></mrow></msub><mo>&times;</mo>
			<msub><mi>p</mi><mi>n</mi></msub><mo>+</mo><mn>1</mn></math>
			</math></p>
			<p>The number <math><mi>q</mi></math> is not divisible by any of the prime numbers
				<math><msub><mi>p</mi><mi>i</mi></msub></math>,
				since it leaves a remainder of 1 when it is divided by any of them. So,
				<math><mi>q</mi></math> must be divisible by some prime that is not in the list,
				by the Fundamental Theorem of Arithmetic. This is a contradiction to our assumption
				of having the complete list of prime numbers. Hence, there must be an infinite
				number of prime numbers.</p>
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MathML HTML

Proof by Contradiction

ContradictionProof.html

Output