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Check out the dot points that say include the words: (gather, process, present, analyse) information from secondary resources, use available evidence

This is for most people right?

Let y = e^x Take the log of both sides and we have ln(y) = ln(e^x) Using the log law ln(m^n) = nln(m), ln(y) = xln(e) Notice that ln(e) = 1 (or using the log law log_a a = 1 for all real a > 0) ln(y) = x If you understand this, then you can try to figure out a quite way of converting y...

Ahhh subtracting 1 (meant to be 1.5) on both sides destroyed my solution About your solution, the last line, when does equality occur for k \in \mathbb{Z}^+?

Alright, fresh start... sorry Beginning from the use of induction hypothesis \geq \frac{2}{3}k\sqrt{k} + \sqrt{k+1} = \frac{2}{3}\sqrt{k+1} \left ( \frac{k\sqrt{k}}{\sqrt{k+1}} +\frac{3}{2}\right ) Now we need to prove that (time for a discovery) \frac{k\sqrt{k}}{\sqrt{k+1}}...

Does dividing both sides by k for this question mean that k=0........? I have just edited my post

Sorry, was a typo

Let predicate P(n) be ".." Basis: For n=1, ... Induction: suppose that k in positive integers Suppose that P(k) is true i.e. √1 + √2 + ... + √k ≥ 2/3 k√k We want to deduce that P(k+1) is true i.e. √1 + √2 + ... + √k + √(k+1) ≥ 2/3 (k+1)√(k+1) We have √1 + √2 + ... + √k + √(k+1) ≥ 2/3...

Question is $Prove by induction that, for n \in \mathbb{Z}^{+}, \\ \centerline{\sqrt{1} + \sqrt{2} + \dots + \sqrt{n} \geq \frac{2}{3}n\sqrt{n}} I'm typing up a super-informal proof right now

Why is 0≥1? I was working on a mathematical induction question and got stuck. I started to work backwards and ended up with this...

q -> p, lol I forgot that, thanks for the revision My method was basically the same thing: $ If $7|x $ and $7|y,$ then there exists positive integers A and B such that$ $ x = 7A and y = 7B. Squaring each and adding together gives \\ x^2 + y^2 = 49(A+B) = 7(M) $($where M = A+B \in \mathbb{Z})...

Yes I realized haha Thanks for all the help! :)

Hello, 2 questions 1. What is that one-line method SpiralFlex? 2. seanig89, is my method now valid? I've fixed some mistakes, please look at my solution (red part)

sorry I made a mistake while finding out the quadratic residues (mod 7) (and another mistake for not checking it with yours and assuming that I'm right)

how about -y^2 cannot be equal to 1,2,4 because there is no solution for y (mod 7)? (removed 3)

Mmm yes :O

Ahhhhhhhhh ok, thank you very much! Yesterday I couldn't sleep because of this question, but I thought of one solution. I've read and understood your solution, but can you please check if my method is valid? :) 7|x^2 + y^2 is the same as saying x^2 +y^2 \equiv 0 \pmod{7} Which is the same...

No idea, but I just had a quick google and saw the words 'congruent modulo'. Now I know where I'm heading (time to use some modulo)

Ahhhh, ok thanks I'll come back when an idea sparks

$ For integers $ x$ and $ y,$ show that $7|x^2 + y^2$ if and only if $7|x$ and $7|y I'm going to prove this, omitting lots of words and formality If 7|x^2 + y^2, then there exists a positive integer M such that x^2 + y^2 = 7M \frac{x^2}{7} + \frac{y^2}{7} = M Since M in an integer, then...