Weekly Challenge 11: Unit interval

7051 /www/nrich/html/content/id/7051/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"> <ul id="buttonBar"> <li> <a href="http://nrich.maths.org/2675&part=">Try this next</a> </li> <li> <a href="http://nrich.maths.org/1382&part=">Read all about it</a> </li> <li> <a href="https://nrich.maths.org/discus/messages/27/27.html?1274115048">Ask NRICH</a> </li> <li> <a href="http://nrich.maths.org/573&part=">Warm-up problem</a> </li> <li> <a href="http://nrich.maths.org/7038&part=solution">Last week's solution</a> </li> </ul> <div> Take any two numbers between $0$ and $1$. Prove that the sum of the numbers is always less than one plus their product. That is, if $0< x< 1$ and $0< y< 1$ then prove $$x+y< 1+xy$$ </div> <div class="framework"> Did you know ... ? Pure inequalities such as this one are often used in the analysis of far more difficult mathematics problems: whilst the inequalities might be simple to prove in themselves, they can be surprisingly useful as tools.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Although this solution is quite short to write down, you do need to keep a clear head to find it! As you read this proof, think carefully about each step to be sure that you follow it. Suppose that $0< x< 1$ and $0< y< 1$ for real numbers $x, y$. Since $0< y< 1$ we must also have $0< 1-y< 1$. Similarly, $0< 1-x< 1$. Since the product of two real numbers between $0$ and $1$ must also be between $0$ and $1$ we have $$0< (1-x)(1-y)=1-x-y+xy< 1$$ Looking at the left hand side of this inequality we have $$ 0< 1-x-y+xy $$ Rearranging gives the desired result. Note: In this proof we assume standard properties of real numbers, such as "the product of two real numbers between $0$ and $1$ must also be between $0$ and $1$". You might wish to read this proof carefully and try to note where assumptions such as these have been made. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Although this solution is quite short to write down, you do need to keep a clear head to find it! As you read this proof, think carefully about each step to be sure that you follow it. Suppose that $0< x< 1$ and $0< y< 1$ for real numbers $x, y$. Since $0< y< 1$ we must also have $0< 1-y< 1$. Similarly, $0< 1-x< 1$. Since the product of two real numbers between $0$ and $1$ must also be between $0$ and $1$ we have $$0< (1-x)(1-y)=1-x-y+xy< 1$$ Looking at the left hand side of this inequality we have $$ 0< 1-x-y+xy $$ Rearranging gives the desired result. Note: In this proof we assume standard properties of real numbers, such as "the product of two real numbers between $0$ and $1$ must also be between $0$ and $1$". You might wish to read this proof carefully and try to note where assumptions such as these have been made. </mdoxml> 2 3 0 0 0 0 1 Weekly Challenge 11: Unit interval A weekly challenge: these are shorter problems aimed at Post-16 students or enthusiastic younger students. Stage 5 Core Mapping Document Polynomials AS Secondary Mapping Document DisplayCabinet