Proof Confusions: How do I approach this?∀y∈R≥0,∀n∈Z+,n&gt;y⇒(∃ϵ∈R≥0,0≤ϵ&lt;1landy=(n+ϵ)2−n2)

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Proof Confusions: How do I approach this?∀y∈R≥0,∀n∈Z+,n&amp;gt;y⇒(∃ϵ∈R≥0,0≤ϵ&amp;lt;1landy=(n+ϵ)2−n2)

amisayq6t · Accepted Answer

Step 1Let&amp;nbsp;y∈R+&amp;nbsp;and&amp;nbsp;n∈N&amp;nbsp;such that&amp;nbsp;n&amp;gt;y. We want to show the existence of&amp;nbsp;ε∈[0,1[&amp;nbsp;&amp;nbsp;such that&amp;nbsp;y=(n+ε)2−n2Let&amp;nbsp;f(ε)=(n+ε)2−n2&amp;nbsp;for all&amp;nbsp;ε∈[0,1[&amp;nbsp;&amp;nbsp;f is a continuous non-decreasing function, thus by the intermediate value theorem,f([0,&amp;nbsp;1[)&amp;nbsp;&amp;nbsp;Butf([0,1[)=[f(0),f(1)[=[0,(n+1)2−n2[=[0,2n+1[n&amp;gt;y&amp;nbsp;so&amp;nbsp;2n+1&amp;gt;y&amp;nbsp;and thus&amp;nbsp;y∈[0,2n+1[&amp;nbsp;&amp;nbsp;and because of what said above, there exists&amp;nbsp;ε∈[0,1[&amp;nbsp;&amp;nbsp;such that&amp;nbsp;y=f(ε)=(n+ε)2−n2

Jonas Dickerson · Accepted Answer

Step 1Solve y=(n+ε)2−n2 sor ε, the equation has two roots ε1=y+n2−n and ε2=−y+n2−n. Then prove that 0≤ε1&amp;lt;1 or 0≤ε2&amp;lt;1. Is easy to show that ε2≤0 and for ε1 note that0≤y&amp;lt;n&amp;lt;2n+1⇒0≤y&amp;lt;2n+1sum n2n2≤y+n2&amp;lt;n2+2n+1⇒n2≤y+n2&amp;lt;(n+1)2take square rootn≤y+n2&amp;lt;(n+1)⇒0≤y+n2−n&amp;lt;1⇒0≤ε1&amp;lt;1

Proof Confusions: How do I approach this? \forall y \in \mathbb{R}^{\geq

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