True story: fairly recently, an old student of mine who became a high school teacher got...
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True story: fairly recently, an old student of mine who became a high school teacher got back in contact with me to ask a math question. He and a fellow high school teacher were arguing about the equation a+b√2=c+d√2 where a, b, c, d are unknown real numbers (this appeared somewhere in the high school math curriculum for some reason). In particular, his colleague thought that one could immediately conclude that a = c and b=d, which seems reasonable, since seemingly one would "equate similar parts" or something like that. However, there is no obvious algebra rule to settle this, and that's why he contacted me. It's an arithmetic/algebra issue for sure, but it is slightly tricky, as it turns out. (a) First of all, as much as we might be tempted to, we're not allowed to just "equate similar parts" in the above expression, as it is possible to have a+b√2 = c+d√2 but NOT have a = c and b=d! So, your task here is to find real number values for a, b, c, and d for which a +b√2 = c+d√2 but it isn't true that a b=d. = c and [Note: yes, this is very different from things we've done previously. But, as it so happens, it will be very similar to things that are coming up very shortly in this course, so it doesn't hurt to get our feet wet now.] [Hint: play with numbers. Try a nice irrational value for one of the unknowns!] (b) Ah, but it does turn out that things behave well for certain types of numbers. So. prove the following statement with a proof by contradiction: If a +b√2=c+d√2 and a, b, c, d are all integers, then actually a = c and b=d! [Hint: setting up the proof by contradiction requires you to simplify ~ (a = c and b = d) with the help of DeMorgan, so first write that carefully. Then notice that there are actually three possibilities (cases) here, when you think about the meaning of the word "or": it might be that ac, b=d; it might be that a = c, b d; finally, it might be that a c, b + d. Find a contradiction in each case!] [Moral of the story; different types of proof techniques combine in unexpected ways! True story: fairly recently, an old student of mine who became a high school teacher got back in contact with me to ask a math question. He and a fellow high school teacher were arguing about the equation a+b√2=c+d√2 where a, b, c, d are unknown real numbers (this appeared somewhere in the high school math curriculum for some reason). In particular, his colleague thought that one could immediately conclude that a = c and b=d, which seems reasonable, since seemingly one would "equate similar parts" or something like that. However, there is no obvious algebra rule to settle this, and that's why he contacted me. It's an arithmetic/algebra issue for sure, but it is slightly tricky, as it turns out. (a) First of all, as much as we might be tempted to, we're not allowed to just "equate similar parts" in the above expression, as it is possible to have a+b√2 = c+d√2 but NOT have a = c and b=d! So, your task here is to find real number values for a, b, c, and d for which a +b√2 = c+d√2 but it isn't true that a b=d. = c and [Note: yes, this is very different from things we've done previously. But, as it so happens, it will be very similar to things that are coming up very shortly in this course, so it doesn't hurt to get our feet wet now.] [Hint: play with numbers. Try a nice irrational value for one of the unknowns!] (b) Ah, but it does turn out that things behave well for certain types of numbers. So. prove the following statement with a proof by contradiction: If a +b√2=c+d√2 and a, b, c, d are all integers, then actually a = c and b=d! [Hint: setting up the proof by contradiction requires you to simplify ~ (a = c and b = d) with the help of DeMorgan, so first write that carefully. Then notice that there are actually three possibilities (cases) here, when you think about the meaning of the word "or": it might be that ac, b=d; it might be that a = c, b d; finally, it might be that a c, b + d. Find a contradiction in each case!] [Moral of the story; different types of proof techniques combine in unexpected ways!
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Organizational Behaviour Concepts Controversies Applications
ISBN: 978-0132310314
6th Canadian Edition
Authors: Nancy Langton, Stephen P. Robbins, Timothy A. Judge, Katherine Breward
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