Algebra for Schools And Colleges
Algebra for Schools And Colleges
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In our eReader you can find the full English version of the book. Read Algebra for Schools And Colleges Online - link to read the book on full screen. Our eReader also allows you to upload and read Pdf, Txt, ePub and fb2 books. In the Mini eReder on the page below you can quickly view all pages of the book - Read Book Algebra for Schools And Colleges
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To quickly assess the difficulty of the text, read a short excerpt:
7 goes into QQ 9 times and 3 over. Hence this means 66 = 7. 9 + 3, or 66-7. 9=1:3.
1 1 DIVISIBILITY OF NUMBERS. 241 333. Peoblem. To find the greatest common divisor of two nuwyhers.
Let m and n be the numbers, and let m be the greater.
1. Divide m by n. If the remainder is zero, n will be the divisor required, because every number divides itself. If there IS a remainder, let q be the quotient and r the remainder.
Then m — nq^ r.
Let d be the common divisor required. Because m and n, are both d...ivisible by ^, m — nq must also be divisible by d (Theorem II). Therefore, ^r is divisible by d. Hence every common divisor of m and n is also a common divisor of n and r. Conversely, because m :=z nq + r, every common divisor of n and r is also a divisor of m. There- fore, the greatest common divisor of m and n is the same as the greatest common divisor o^ n and r, and we proceed with these last two numbers as we did with in and n, 2. Let r go into n q' times with the remainder t\ Then n = rq' -{- r\ 'Or n ^rq' — /.
What to read after Algebra for Schools And Colleges?
You can find similar books in the "Read Also" column, or choose other free books by Newcomb Simon to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
7 goes into QQ 9 times and 3 over. Hence this means 66 = 7. 9 + 3, or 66-7. 9=1:3.
1 1 DIVISIBILITY OF NUMBERS. 241 333. Peoblem. To find the greatest common divisor of two nuwyhers.
Let m and n be the numbers, and let m be the greater.
1. Divide m by n. If the remainder is zero, n will be the divisor required, because every number divides itself. If there IS a remainder, let q be the quotient and r the remainder.
Then m — nq^ r.
Let d be the common divisor required. Because m and n, are both d...ivisible by ^, m — nq must also be divisible by d (Theorem II). Therefore, ^r is divisible by d. Hence every common divisor of m and n is also a common divisor of n and r. Conversely, because m :=z nq + r, every common divisor of n and r is also a divisor of m. There- fore, the greatest common divisor of m and n is the same as the greatest common divisor o^ n and r, and we proceed with these last two numbers as we did with in and n, 2. Let r go into n q' times with the remainder t\ Then n = rq' -{- r\ 'Or n ^rq' — /.
What to read after Algebra for Schools And Colleges?
You can find similar books in the "Read Also" column, or choose other free books by Newcomb Simon to read onlineMoreLess
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