Prove that √2-1 1 1 √2-1 and hence the two rectangles are similar *Figure 1.18* Deduce that if a rectangle with long side a and short side b has the same shape as the two in the figure 1.13, then so has the rectangle with long side b and short side a-2b (Hint it should look similar to a proof for the irrationality of the square root of two. n (m-2n) Suppose that √√2 + 1 = m² when m and n are natural numbers with m as small as possible, deduce that n √2+1= show this is a contradiction and why.
Prove that √2-1 1 1 √2-1 and hence the two rectangles are similar *Figure 1.18* Deduce that if a rectangle with long side a and short side b has the same shape as the two in the figure 1.13, then so has the rectangle with long side b and short side a-2b (Hint it should look similar to a proof for the irrationality of the square root of two. n (m-2n) Suppose that √√2 + 1 = m² when m and n are natural numbers with m as small as possible, deduce that n √2+1= show this is a contradiction and why.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:The image displays Figure 1.18, titled "A pair of similar rectangles."
This diagram illustrates two rectangles, both of which have one side labeled as "1." The other dimensions differ in length: the larger rectangle has a length labeled as "√2 + 1," while the smaller rectangle's length is "√2 - 1."
The two rectangles are positioned side by side, emphasizing their similar shape but different sizes, which is a key property of similar figures in geometry.
![**Transcription for Educational Website:**
Prove that
\[
\frac{\sqrt{2} - 1}{1} = \frac{1}{\sqrt{2} - 1}
\]
and hence the two rectangles are similar *Figure 1.18*.
Deduce that if a rectangle with long side \(a\) and short side \(b\) has the same shape as the two in the figure 1.13, then so has the rectangle with long side \(b\) and short side \(a - 2b\) (Hint: it should look similar to a proof for the irrationality of the square root of two).
Suppose that \(\sqrt{2} + 1 = \frac{m}{n}\) when \(m\) and \(n\) are natural numbers with \(m\) as small as possible, deduce that
\[
\sqrt{2} + 1 = \frac{n}{(m - 2n)}
\]
show this is a contradiction and why.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe168234c-8641-43ba-b22b-28f74f97e6ae%2F262a4de9-114e-4cfb-b653-59a562159cd7%2F18bh80f_processed.png&w=3840&q=75)
Transcribed Image Text:**Transcription for Educational Website:**
Prove that
\[
\frac{\sqrt{2} - 1}{1} = \frac{1}{\sqrt{2} - 1}
\]
and hence the two rectangles are similar *Figure 1.18*.
Deduce that if a rectangle with long side \(a\) and short side \(b\) has the same shape as the two in the figure 1.13, then so has the rectangle with long side \(b\) and short side \(a - 2b\) (Hint: it should look similar to a proof for the irrationality of the square root of two).
Suppose that \(\sqrt{2} + 1 = \frac{m}{n}\) when \(m\) and \(n\) are natural numbers with \(m\) as small as possible, deduce that
\[
\sqrt{2} + 1 = \frac{n}{(m - 2n)}
\]
show this is a contradiction and why.
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