Proofs Inequality Question (1 Viewer)

hmim · Apr 1, 2023

Can someone show their working out for 8 ii) please

Masaken · Apr 1, 2023

oh hey, i think i've done this handout lol
basically apply the expression you proved in part (i) then later the given expression in the q

hmim · Apr 2, 2023

Masaken said:
oh hey, i think i've done this handout lol
basically apply the expression you proved in part (i) then later the given expression in the qView attachment 38107

Thank you so much !

hmim · Apr 2, 2023

Hey Would someone be able to show their working out for this question please?

hmim · Apr 2, 2023

Could someone show how they would work out this question, please?

hmim · Apr 2, 2023

Can someone show their working out for this question please?

axyx · Apr 2, 2023

$(ay-bx)^2 \ge 0$
$a^2 y^2 -2aybx+b^2 x^2 \ge 0$
$a^2 y^2 + b^2 x^2 \ge 2aybx$
$-a^2 y^2 - b^2 x^2 \le -2aybx$
$-a^2 y^2 - b^2 x^2 +(a^2 x^2 +b^2 y^2) \le -2aybx+(a^2 x^2 +b^2 y^2)$
$a^2 x^2 -a^2 y^2 - b^2 x^2 +b^2 y^2 \le a^2 x^2 -2aybx+b^2 y^2$
$(a^2 -b^2)(x^2 -y^2) \le (ax-by)^2$

axyx · Apr 2, 2023

$y^2 >0, x^2 >0$
$(x^2 +y^2) >0$ *
$(x-y)^2 >0$
$2(x-y)^2 >0$
$2x^2 -4xy+2y^2 >0$
$2x^2 -4xy+2y^2 +(x^2 +y^2)>(x^2 +y^2)$
$3x^2 -4xy+3y^2 >(x^2 +y^2)$
From *:
$(x^2 +y^2) >0$
$\therefore 3x^2 -4xy+3y^2 >0$

hmim · Apr 3, 2023

h

tywebb said:
$1+3+2+6=12\text{ is divisible by }12\therefore\text{ it is true for }n=0$ .

$\text{If it is true for }n=k\text{ then }35^k+3\times7^k+2\times5^k+6=12M\text{ for an integer }M$

$\begin{aligned}\therefore35^{k+1}+3\times7^{k+1}+2\times5^{k+1}+6&=35\times35^k+21\times7^k+10\times5^k+6\\&=(36-1)\times35^k+(24-3)\times7^k+(12-2)\times5^k+12-6\\&=12(3\times35^k+2\times7^k+5^k+1)-35^k-3\times7^k-2\times5^k-6\\&=12(3\times35^k+2\times7^k+5^k+1)-12M\\&=12(3\times35^k+2\times7^k+5^k+1-M)\end{aligned}$
$\text{ is divisible by }12\text{ since }3\times35^k+2\times7^k+5^k+1-M\text{ is an integer}$
$\text{and so it is true for }n=k+1$ .

$\text{Hence by the principal of mathematical induction it is true for all integers }$
$n\ge0$ .

Thanks!

hmim · Apr 3, 2023

axyx said:
$y^2 >0, x^2 >0$
$(x^2 +y^2) >0$ *
$(x-y)^2 >0$
$2(x-y)^2 >0$
$2x^2 -4xy+2y^2 >0$
$2x^2 -4xy+2y^2 +(x^2 +y^2)>(x^2 +y^2)$
$3x^2 -4xy+3y^2 >(x^2 +y^2)$
From *:
$(x^2 +y^2) >0$
$\therefore 3x^2 -4xy+3y^2 >0$

Thanks!

hmim · Apr 3, 2023

tywebb said:
As with many divisibility proofs however it is probably better to do it without induction.

$35^n+3\times7^n+2\times5^n+6=(7^n+2)(5^n+3)$

$\begin{aligned}5^n+3&=5^n-1+4\\&=(5-1)(5^{n-1}+5^{n-2}+\cdots+1)+4\\&=4(5^{n-1}+5^{n-2}+\cdots+1+1)\text{ is divisible by 4 and}\end{aligned}$

$\begin{aligned}7^n+2&=7^n-1+3\\&=(7-1)(7^{n-1}+7^{n-2}+\cdots+1)+3\\&=3(2(7^{n-1}+7^{n-2}+\cdots+1)+1)\text{ is divisible by }3\end{aligned}$

$\therefore 35^n+3\times7^n+2\times5^n+6\text{ is divisible by }12$ .

Thanks!

hmim · Apr 3, 2023

Can someone show me their working for this question please?

hmim · Apr 3, 2023

Can someone show me their working out for ii) please?

hmim · Apr 3, 2023

Can someone show me their working to the above question, please?

hmim · Apr 3, 2023

Can someone show me their working out to this q please?

jimmysmith560 · Apr 3, 2023

Would the following working help?

Part (i):

Part (ii):

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Proofs Inequality Question (1 Viewer)

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