 | James Wood - 1815 - 338 σελίδες
...irrational part. Thus, a^/x±b^/x = a±b.^/lc', ,^^±5^3 = 10«/3 ±5.>/3 = l5*/3, or 5^/3". (252.) The square root of a quantity cannot be partly rational and partly a quadratic surd. If possible, let */n=a + +/m; then by squaring both sides, n = a* + 2a*^/m + m, and 2a^m = n — a*—... | |
 | James Ryan, Robert Adrain - 1824 - 542 σελίδες
...4-^/72 to a universal surd. Ans. .J/9. } V. METHOD OF EXTRACTING THE SQUARE ROOT OF BINOMIAL SURDS. 363. The square root of a quantity cannot be partly rational and partly a quadratic surd. If possible, let ; tnen by squaring both sides, n=aa + , and 2a^/m=n — a2 — m ; therefore, v/nz =... | |
 | Charles Tayler - 1824 - 350 σελίδες
...is dependent, and by which it is in a great measure guided. These theorems are as follow, viz. 373. The square root of a quantity cannot be partly rational, and partly a quadratic surd. If it be possible, let */x—a± */b, where a is a rational quantity, and */b an irreducible surd. We... | |
 | James Ryan - 1826 - 430 σελίδες
...$/72 to a universal surd. Ans. § V. METHOD OF EXTRACTING THK SQUARE ROOT OF BINOMIAL SURDS. 363. Tiie square root of a quantity cannot be partly rational and partly a quadratic surd. If possible, let A/»=a then by squaring both sides, n=«2+2a^//n+m, and n — aa — m ; therefore, </m=... | |
 | John Bonnycastle - 1829 - 372 σελίδες
...imaginary surd, the same theorems will still hold, by only changing— 6 into +b, as below. * Prop. 1. The square root of a quantity cannot be partly rational and partly a quadratic surd. If possible, let ^/ n~a-^-^/m ;then by squaring both sides, » = oa +2nv/mm, and by transposition, Sov^m... | |
 | John Bayley (fellow of Emmanuel College, Cambridge.) - 1830 - 428 σελίδες
...number consists of (3м) or (3и— 1) or (3и— 2) digits, when its cube root consists of (ri) digits. The square root of a quantity cannot be partly rational and partly irrational ; for, if possible, let \/a — b + f- j- a — V — с . • . a— V + 2i \/c + с .... | |
 | 1832 - 410 σελίδες
...progression, the first term is to the third, as the square of the first is to the square of the second. 5. The square root of a quantity cannot be partly rational and partly a quadratic surd. 6. Prove the Binomial theorem: — 1st, when the index is a whole positive number ; 2d, when the index... | |
 | John Radford Young - 1832 - 408 σελίδες
...susceptible of the requisite extraction ; and therefore cannot be otherwise accurately expressed. THEOREM 1. The square root of a quantity cannot be partly rational and partly a quadratic surd. For if -y/ a = 6 + Vc, then, by squaring each side, we shall have a = 63 + 26 \/c + c, and 26-/ с... | |
 | James Ryan, Robert Adrain - 1835 - 388 σελίδες
...3{/$+{/72 to a universal surd. Ans. § V. METHOD OF EXTRACTING THE SQUARE ROOT OF BINOMIAL SURDS. 300. The square root of a quantity cannot be partly rational and partly a quadratic surd. If possible, let ^/n=a-\-^/m ; then by squaring both sides, n=a?-^-'2a^/m-{-m, and 2a*jm=n — a" —... | |
 | James Bryce - 1837 - 322 σελίδες
...quantities ; but are rational í also, 4/(2a), ^(аг+хг), &c. are surds.* PROPOSITIONS. 112. PROP. I. The square root of a quantity cannot be partly rational and partly irrational. For, if possible, let ^/a = 6+^/c; then (Art. 31) а = 6г+26,/с + с, and (Art. 28) а... | |
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Tiie square root of a quantity cannot be partly rational and partly a quadratic surd. If possible, let... 
